Modern Physics for IITJEE

Modern Physics IIT–JEE Syllabus: Alpha, beta and gamma radiations, law of radioactive decay, decay constant, half-life and mean life. Photoelectric effect. De Broglie wavelength. Bohr’s theory of hydrogen like atoms. Production of characteristic and continuous X-rays, properties of X-rays. Atomic nucleus, binding energy and its calculation. Fission and fusion processes, energy calculations in these processes.

1.

STRUCTURE OF ATOM

All matter is made up of tiny particles known as atoms. There are only about 105 different kinds of atoms, and they combine with each other in different ways to form groups called molecules. All matter has been found to be composed of atoms or molecules, and some knowledge of how atoms are made will give us valuable information about the behaviour of matter.

Thomson's Atomic Model On the basis of his experiments J.J. Thomson proposed a model of internal atomic structure according to which atom consisted of positively charged substance (+ve electric fluid) distributed uniformly over the entire body of the atom, with negative electrons embedded in this continuous positive charge like seeds in a watermelon. It was a good effort to reveal mystery of an atom but it was not the true picture of an atom.

Rutherford's Atomic Model The correct description of the distribution of positive and negative charges within an atom was made in 1911 by a New Zealander when working at Manchester University in England. This was Ernest Rutherford, who was later made Lord Rutherford for his many scientific achievements. He entered into physics during that crucial period of its development when the phenomenon of natural radioactivity had just been discovered, and he was first to realize that radioactivity represents a spontaneous disintegration of heavy unstable atoms. Rutherford realized that important information about the inner structure of atoms could be obtained by the study of collisions between on rushing  particles and the atoms of various materials forming the target. Collimator

Zinc Sulphate Screen

Microscope



Particle Source

Gold Foil

Fig.(1) Schematic diagram of the experimental set up used by Rutherford.

 The basic idea of the experimental arrangement used by Rutherford in his studies was explained as follows: a speck of  - emitting radioactive material; a lead shield with a hole that allowed a narrow beam of the   particles to pass through; a thin metal foil to deflect or scatter them; and a pivoted flourescent screen with a magnifier through which the tiny flashes of light were observed whenever an -particle struck the screen. The whole apparatus was evacuated, so that the particles would not collide with air molecules. Observation Most of the -particles penetrated the foil with very little deflection. An appreciable fraction of them were deflected through large angles - a few were turned back almost as though they had been reflected from the foil. This was a deflection of nearly 180° and a completely impossible phenomenon according to the Thomson's model. Nucleus Such large deflections required strong forces to be acting, such as those between very smaller charged particles very close together. This would be possible, Rutherford reasoned, if all the Fig.(2) According to Rutherford’s nuclear model, the alpha positive charge, along with most of the atomic particles were scattered by the coulomb force of a tiny particle (the nucleus) rather than a large mass, were concentrated in a very small central sphere, as in Thomson’s model of the atom, Each region which Rutherford called the atomic alpha particle experienced a single string collision. nucleus. Rutherford, knowing the kinetic energy of the -particles, calculated that they would be within about 10-12cm from its centre if  - particles were to be turned back in the direction from which they came. Because there would be a Coulomb force of attraction between the positive nucleus and the negative electrons, the two would be down together and the atom would vanish unless some provisions were made to prevent it. It was suggested that the electrons might be orbiting rapidly around the nucleus, so that the electrostatic attraction would merely provide the necessary centripetal force.

Drawbacks (i) Rutherford's atomic model was unable to make any predictions about the light that an atom would emit (ii) More serious than this was its conflict with the accepted laws of electromagnetic theory. An electron revolving rapidly around a nucleus must have a continual centripetal acceleration, and this acceleration would cause a continuous loss of energy by radiation. Bohr calculated that this emission of radiation would cause the electrons in an atom to lose all their energy and fall into the nucleus within a hundred - millionth of a second. Since matter composed of atoms exists permanently, as far as we know, there was obviously something wrong here. Bohr's conclusion was that the conventional classical laws of physics must be wrong, at least when applied to the motion of electrons within an atom.



Bohr’s Theory Bohr in defiance of the well - established laws of classical mechanics and electrodynamics, proposed that the following rules must hold 1. Of all the infinite number of mechanically possible orbits for an electron revolving around a nucleus, only a few are permitted. These are the orbits in which the angular momentum of the electron is an integral multiple of h/2.

v

mv 2 r

Fig.(3)

Planetary model of the hydrogen atom. A massive positive charge is nearly stationary in the center. An electron travels in a circular orbit. The electron is held in its orbit by the Coulomb’s law attraction of the positive nucleus and the negative electron.

2. While circling around these permitted orbits, the electrons do not emit any electromagnetic radiation, even though conventional electrodynamics holds that they should. 3. Electrons may jump from one orbit to another, in which case the difference in energy between the two states of motion is radiated as a photon whose frequency is determined by the quantum rule E = hf. Bohr's Orbits

For an electron orbiting in a hydrogen atom, the necessary centripetal force is the electrostatic attraction between the negative electron and the massive, positively-charged proton, that is the nucleus. e2 ke2 mv 2 Thus, k 2 = or r= (1) r r mv 2 According to Bohr's theory nh mvr = where, n = 1, 2, 3,…. 2 nh r= (2) 2mv From (1) and (2) 2ke2 v= nh r

n 2h 2 4 kme 2

2

where h = 6.63  10-34 J –s

(1)



Electron energies Kinetic energy

E 2 2

2 k me 1 2 m  2ke  mv = = (2) 2 2  nh  n 2h 2 Potential energy  4 2 kme 2   4 2k 2 me4 ke 2 = U= = ke2   n2h2  r n 2h 2  

K=

2 2

4

Free Electron

0 eV

-3.40eV

n = 5 Bound Electron n=4 n=3 n=2

-13.6eV

n=1

-1.51eV

(3) Total energy

1  2 2 k 2me 4   n 2  h2  9 Putting the values of k = 9  10 Nm2/C2 e = 1.6  10-19 C and h = 6.63  10-34 Js, we get 1  13.6 E =  2 (2.18  10-18) J = eV n n2 E=K+U=

(4)

(5)

Fig.(4) Energy level diagram for the Bohr model of the hydrogen atom. The vertical axis represents energy. The (arbitrary) zero of energy is taken as the energy of a stationary electron, infinitely far from the positive nucleus. The lowest energy level (n = 1) is known as the ground state.

Radiation and Energy Levels E = hf E = E2  E1 Using equation (5)  1 1  E = 2.18  10-18  2  2  n   1 n2  Applying Planck’s Law,  1 E 1  f= = 3.29  1015  2  2  Hz n  h  1 n2  Dividing the above equation by c = 3  108 m/s, we get 1 3.29  1015  1 1  -1     m  3  10 8  n12 n 22  1 1 1  R  2  2  m-1   n1 n2  where R = Rydberg constant = 1.097  107 m-1 .

or

(6)

(7)



0 eV Bracket

-1.51eV -3.40eV

-13.6 eV

Paschen Balmer

Lyman

n= n=5 n=4 n=3 n=2

n=1

Fig.(5) Light is emitted from the hydrogen atom only when the electron makes transitions between stationary orbits. The Balmer series of spectral lines, for instance, results when electrons from higher energy levels fall into the n = 2 level, releasing their energy as a single photon.

Successes and Limitations Bohr showed that Planck's quantum idea were a necessary part of the atom and its inner mechanism; he introduced the idea of quantized energy levels and explained the emission or absorption of radiation as being due to the transition of an electron from one level to another. As a model for even multielectron atoms, the Bohr picture is still useful. It leads to a good, simple, rational ordering of the electrons in larger atoms and qualitatively helps to predict a great deal about chemical behaviour and spectral details.

Bohr's theory is unable to explain the following facts 1. The spectral lines of hydrogen atom is not a single line but a collection of several lines very close together. 2. The structure of multielectron atoms is not explained. 3. No explanation for using the principle of quantisation of angular momentum. 4. No explanation for Zeeman effect If a substance which gives a line emission spectrum is placed in a magnetic field, the lines of the spectrum get split up into a number of closely spaced lines. This phenomenon is known as Zeeman effect.

Conclusion The atom consists of a heavy positively charged nucleus and negatively charged electrons moving around it. The electron is an elementary particle having a mass me  9.1  1031 kg and a charge e, e being an elementary charge approximately equal to 1.60  10-19C.

 The nuclear charge is equal to +Ze, where Z is the atomic number. The atom contains Z electrons, their total charge being Ze. Consequently, the atom is an electrically neutral system. The size of the nucleus varies depending on Z from 10-13 cm to 10-12 cm. The size of the atom is a quantity of the order of 10-8 cm. The energy of the atom is quantized. This means that it can assume only discrete (i.e. separated by finite gaps) values: E1, E2, E3,…, which are called the energy levels of the atom (E1 < E2 < E3 < …). Atoms with different Z's have different sets of energy levels. In a normal (unexcited) state, the atom is on the lowest possible energy level. In such a state, the atom may stay for an infinitely long time. By imparting an energy to the atom, it is possible to transfer it to an excited state with an energy higher than the energy of the ground state. A transition of the atom to a higher energy level may occur as a result of absorption of a photon or as a result of a collision with another atom or a particle, say, an electron. Excited states of the atom are unstable. The atom can stay in an excited state for about 10-8 s. After that the atom spontaneously (by its own) goes over to a lower energy level, emitting in this process a photon with an energy E ik  hf ik (i > k), where i is the number of the energy level in the initial state and k is the number of the level to which a spontaneous transition of the atom occurred. For example, an atom which is in an excited state with the energy E3 can return to the ground state either directly, by emitting a photon of frequency f31 = (E3  E1)/h, or through an intermediate state with the energy E2, as a result of which two photons with frequencies f32 = (E3  E2)/h and f21 = (E2  E1)/h are emitted.

Important Formulae 1.

Radius of nth orbit n2 rn = 0.53 Å where Z = atomic number (8) Z

2.

Velocity of the electron in the nth orbit Z c  8 vn =   where c = 3  10 m/s n  137 

3.

Energy of the electron in the nth orbit Z2 En = -13.6 2 (eV) n Z2 En = -(2.18  10-18) 2 (J) n E=K+U U K = -E =  2 U = 2E = -2 K

(9)

(10 a) (10 b) (11 a ) (11 b) (11 c)

 4.

Wavelength of photon emitted for a transition from n2 to n1 1 1 1 (12)  R Z 2  2  2    n1 n2  where R = 1.096  107 m-1 (for a stationary nucleus) If nucleus is not considered to be stationary R R= (13) m 1 M where m is the mass of electron and M is the mass of nucleus.

5.

Wavelength (Å) of a photon of energy E (eV) is given by 12400 = Å (14) E (eV)

6.

Momentum of a photon of energy E E p= (15) c Example: 1 A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the third Bohr orbit. (a) Find the value of Z (b) Find the energy required to excite the electron from n = 3 to n = 4 (c) Find the wavelength of radiation required to remove electron from first Bohr’s Orbit to infinity. (d) Find the kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit. Solution (a) Given E23 = 47.2 eV  1 1  Since E = 13.6Z2  2  2  eV n   1 n2  1  1  47.2 = 13.6 Z2  2  2   Z=5 3  2 (b) To find E34; n1 = 3; n2 = 4   1 1  E = 13.6 Z2  2  2  eV n   1 n2  1   1  E = 13.6  52  2  2  = 16.53 eV 4  3 (c) Ionization energy is the energy required to excite the electron from n = 1 to n =  1  1 Thus, E = 13.6  52  2  2  = 340 eV   1

 The respective wavelength is hc 12400 12400 = = = 36.47 Å  E E 340 (d) K = -E = +340 eV U = 2E = -680 eV h 6.63  10 34 L= = = 1.056  10-34 J-s 2 2 Example 2 Find the quantum number n corresponding to excited state of He+ ion if on transition to the ground state, the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm. The ionization energy of H atom is 13.6 eV. Solution The energy transitions for the given wavelengths are 12400 12400 E1 =   11.43eV 1 1085 12400 12400 E2 = = 40.79 eV  304 2 Total energy emitted E = E1 + E2 = 52.22 eV   1 1  Now E = 13.6 Z2  2  2  eV E = energy emitted n  n 2   1 1 1   or 52.34 = 13.6  22  2  2  n  1 Thus, n = 5 Example 3 An isolated hydrogen atom emits a photon of 10.2 eV. (a) Determine the momentum of photon emitted (b) Calculate the recoil momentum of the atom (c) Find the kinetic energy of the recoil atom. [Mass of proton, mP = 1.67  10-27 kg]

Solution (a) Momentum of the photon is E 10.2  1.6  10 19 p1 = = = 5.44  10-27 kg m/s c 3  10 8 (b) Applying the momentum conservation p2 atm

photon

p1

p2 = p1 = 5.44  10-27 kg m/s (c) K =

1 2 mv (v = recoil speed of atom, m = mass of hydrogen atom) 2

 2

1  p p2 m   2 m 2m Substituting the value of the momentum of atom, we get or

K=

5.44  10 

27 2

K=

2  1.67  10

 27

= 8.86  10-27 J

Example 4 A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV. Find the energy and wavelength of photon emitted. Solution Since the binding energy is always negative, therefore, Ei = -0.85 eV Let ni be the initial binding state of the electron, then Z2 En =  13.6 ni2 or

-0.85 = -13.6

Z2 ni2

or ni = 4 Binding energy = En = -13.6 Z2/n2  0.85 eV = -13.6(1)2/n22  n2 = 4 Let the electron now goes to an energy level n whose excitation energy is 10.2 eV. Since the excitation energy E is defined with respect to ground state, therefore  1 1  E = 13.6 Z2  2  2  eV n   1 n2  1 1  or 10.2 = 13.6  12  2  2  1 n f   thus nf = 2 So the electron makes a transition from energy level ni = 4 to nf = 2. Thus, the energy released is E = E4 – E2 1 1 or E = 13.6  2  2  = 2.55 eV 4  2 hc 12400 Since  = =  5511 Å E 2.25eV Example 5 A particle of charge equal to that of an electron, -e and mass 208 times the mass of electron (called a -meson) moves in a circular orbit around a nucleus of charge +3e (take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system: (i) Calculate the radius of nth Bohr orbit (ii) Find the value of n, for which the radius of orbit is approximately the same as that of first Bohr orbit for the hydrogen atom;

 (iii)Find the wavelength of radiation emitted when the -meson jumps from the third orbit to first orbit (Rydberg’s constant = 1.097  107/m) Solution (i) Radius of the nth Bohr orbit for hydrogen atom is n2 rn = 0.53 Z 1 Since r  m Radius of nth orbit for -meson is 0.53n 2 rn = 208Z or rn = (8.5  10-4)n2 (ii) (8.5  10-4)n2 = 0.53  n2 = 623 or n  25 (iii)In case of hydrogen like atom, 1   E = E3 – E1 = 13.6 Z2 1  2  = 12.08 eV  3  since E  m  -meson, E = (12.8)(208) = 22.6 keV 12400 12400 Thus  =  = 0.548 Å E 22.6  103

2.

de BROGLIE WAVES

The first step in the understanding of the hidden meaning of Bohr's quantum orbits was made by a Frenchman, Louis de Broglie, who tried to draw an analogy between the sets of discrete energy levels that characterise the inner state of atoms and the discrete sets of mechanical vibrations that are observed in the case of violin strings, organ pipes etc. de Broglie asked himself, "Could it not be that the optical properties of atoms are due to some kind of standing waves enclosed within themselves?" As a result of these considerations, de Broglie came out with his hypothesis that the motion of electron within the atom is associated with a peculiar kind of waves which he called "pilot waves". In order to have n complete wavelengths (n) fit into the circumference of the nth orbit, the following relation must be true: nn = 2rn From Bohr's theory of the hydrogen atom, n 2h 2 rn = 4 2kme2



or

nn =

or

n =

n 2h 2 2kme2 nh 2

2kme2 h n  mvn

=

nh h m 2ke2

(16)

The wavelength of the wave associated with a moving particle is equal to Planck's quantum constant divided by the momentum of the particle. Example 6 What is the energy and wavelength of a thermal neutron ? Solution By definition, a thermal neutron is a free neutron in a neutron gas at about 20o C (293 K). Since it has three degrees of freedom, therefore 3 3 K = kT  1.38  10  23 293  6.07  10 21 J 2 2 6.63  1034 h h  =  = 0.147 nm p 2mo K 2 1.67  10 27 6.07  10 21











Example 7 A particle of mass m is confined to a narrow tube of length L. (a) Find the wavelengths of the de-Broglie waves which will resonate in the tube, (b) Calculate the corresponding particle momenta, and (c) Calculate the corresponding energies. Solution (a) The de Broglie waves will resonate with a node at each end of the tube. A few of the possible resonance forms are as follows : 2L n = ; n = 1, 2, 3, ...... n (b) Since de-Broglie wavelengths are h n = pn h nh  pn =  n = 1, 2, 3.... n 2 L (c) The kinetic energies of the particles are p 2 n 2h 2 Kn = n  2 , 2m 8 L m n = 1, 2, 3, ........

L N

A

N

N

A N

N

A

A

N

L = (/2)

N

L = 2(/2)

N

L = 3(/2)

N A

A



3.

X – RAYS

Intensity

When a heavy metal target is bombarded with high-energy (30 – 50keV) electrons, it emits X rays. The radiation involves both a continuous and a line spectrum, as shown in the fig.(6). The continuous spectrum, which starts at some minimum wavelength o, arises from the rapid deceleration of the electrons when they enter the target – it is called bremsstrahlung, or braking radiation. The existence of minimum wavelength (or maximum frequency) is further evidence in favour of the photon concept. The highest frequency photon is emitted when an electron loses all its energy in one step. By equating the energy of the electron (eV) to the energy of the photon (hfo), we find hfo = eV hc or o = (17) eV

min

 

 

Fig.(6) The X-ray emission when stream of fast moving electrons strike a target of heavy element.

The minimum wavelength depends on the electron energy, but not on the target material.

Intensity

Intensity W

K3

Mo K2

Cr O

 

K1  3

O

Wavelength

Fig.(7) Different targets of tungsten (W), Molybdenum (Mo) and Chromium (Cr) are used and the kinetic energies of the incident electrons are kept constant. It is observed that the minimum wavelength o is independent of target material.

Fig.(8)

 2

 1

Wavelength

Electron beam of different energies K1, K2 and K3 are incident on a target of same material. The minimum wavelength is inversely proportional to the kinetic energy. If K3 > K2 > K1 , then 3 < 2 < 1. 

N M

L

K K

K

L

L

K

Fig.(9) Energy level diagram for an electron in an atom. The arrows indicate the transitions that give rise to the different series of X – rays.

f

O

Atomic number Z

Fig.(10) A plot of the square root of the frequency versus atomic number Z.

 The line spectrum depends on the element used as target. These characteristic X rays are produced when an electron knocks out an atomic electron from one of the inner levels. The ejected electron leaves a vacancy, which is then filled by an electron falling from a higher level. In the process a high – energy photon is emitted. If the transitions are to the n = 1 level, the X rays are labeled K, K…….If they are to the n = 2 level, they are labeled L, L… In 1913, Moseley noted that the characteristic lines shifted systematically as the target material was changed. He plotted the square root of the frequency of the K line versus the atomic number Z for many elements. The straight line he obtained is shown in the fig.(10). Moseley’s plot did not pass through the origin. Let us see, why?. Once one of the two electrons in the n = 1 level is ejected, an electron in the next highest level will drop to the lower state to fill the vacancy and in the process it emits the K frequency. For this electron the electric field due to the nucleus is screened by the remaining electron in the n = 1 level. Moseley estimated that the effective nuclear charge for the K transition is (Z – 1)e. Thus Moseley’s law for the frequency of the K line is f K  = a(Z – 1) (18)

3 RC where R is the Rydberg constant c is the speed of light. 4 The wavelength of K – lines is given by 1 1   Z  12 1  2  where n = 2, 3, 4,………… (19)   n  when a =

Example 8 Find the cut-off wavelength of the X-rays emitted by an X-ray tube operating at 30 kV. Solution For minimum wavelength, the total kinetic energy should be converted into an X-ray photon. Thus, 12400 hc 12400 =   = 0.41 Å E E 30  103 Example 9 Show that the frequency of K X-ray of a material equals to the sum of frequencies of K and L X-rays of the same material. Solution M

L

L K

K K

The energy level diagram of an atom with one electron knocked out is shown above. Energy of K X-ray is EK = EL - EK of K X-ray is EKβ  EM - EK

 and, thus, or

4.

of L X-ray is EL = EM - EL EKβ  EK + EL f K   f K  f L

RADIOACTIVITY

Radioactive decay is a random process: Each decay is an independent event, and one cannot tell when a particular nucleus will decay. When a given nucleus decays, it is tranformed another nuclide, which may or may not be radioactive. When there is a very large number of nuclei in a sample, the rate of decay is proportional to the number of nuclei, N, that are present dN (20)   N dt Where  is called the decay constant. This equation may be expressed in the form

dN  dt N

and integrated N

dN

t

   dt  No N 0 to yield

 N    t ln  No  where No is the initial number of parent nuclei at t = 0. The number that survive at time t is therefore N = Noe-t (21) This function is plotted in Fig.(11) N No

0.5 No 0.37 No

T1/2 1/

t

Fig.(11)The number of radioactive nuclei in a sample as a function of time. The half-life is the time required for the number to fall to 50% of any initial value (not just t = 0).

The time required for the number of parent nuclei to fall to 50% is called the half-life, T, and may be related to  as follows. Since 0.5 No = No e T

  we have T = ln|2| = 0.693. Therefore 0.693 T= (22)  It takes one half-life to drop to 50% of any starting value. The half-life for the decay of the free neutron is 12.8 min. Other half-lives range from about 10-20 s to 1016 years. Since the number of atoms is not directly measurable, we measure the decay rate or activity (A) dN A=. On taking the derivative of equation we find dt A = N = Noe-t = Roe-t (23) where A = N is the initial activity. The activity is characterized by the same half-life. The SI unit for the activity is the becquerel (Bq), but the curie (Ci) is often used in practice. 1 becquerel (Bq) = 1 disintegration per second (dps) 1 curie = 3.7  1010 dps 1 rutherford = 106 dps Mean life of a radioactive sample is defined as the average of the lives of all nucleus.



 N oe

Tav =

 t

0

No

dt 

T 1   0.693

(24)

Example 10 The half-life of Cobalt - 60 is 5.25 years. How long after its activity have decreased to about one-eigth of its original value ? Solution The activity is proportional to the number of undecayed atoms. In each half-life, half the remaining sample decays.  1  1  1 1 Since        , therefore, three half-lives or 15.75 years are required for the  2   2   2  8` 1 sample to decay to th its original strength. 8 Example 11 A count rate-meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. (a) Find the decay constant (b) Also, find the half life of the sample Solution dN Initial velocity Ai =  N o  4750 (i) dt t  0 Final velocity

Af =

dN dt

 N  2700

(ii)

t 5

Dividing (i) by (ii), we get 4750 N o  2700 Nt

(iii)

 The decay constant is given by 2.303 N = log o t Nt 2.303 4750 or = = 0.113 min-1 log 5 2700 Half life of the sample is 0.693 0.693 T=   6.14 min  0.113 Example 12 The mean lives of a radio active substance are 1600 and 400 years for  - emission and  - emission respectively. Find out the time during which three fourth of a sample will decay if it is decaying both by  - emission and  - emission simultaneously. Solution When an substance decays by  and  emission simultaneously, the average disintegration constant av is given by av =  +  where  = disintegration constant for  - emission only  = disintegration constant for  - emission only Mean life is given by Tm = 1/ 1 1 1 1 1  av =  +  or = 3.12  10-3     Tm T T 1600 400 avt = 2.303 log

No Nt



(3.12  10-3)t = 2.303 log



t = 2.303 

100 25

1 log 4 = 443.5 years 3.12  10 3

There are two types of radioactivity, natural radioactivity observed in unstable elements in nature and artificial radioactivity observed in artificially obtained isotopes. Example 13 The half-life of radium is 1620 years. How many radium atoms decay in 1s in a 1g sample of radium. The atomic weight of radium is 226 kg/mol. Solution Number of atoms in 1 g sample is  0.001 26 21 N=   6.02  10  2.66  10 atoms.  226  The decay constant is 0.693 0.693 =   1.35  1011 s-1 7 T1 / 2 1620 3.16  10





Taking one year = 3.16  10 s 7













N  N  1.35  1011 2.66  10 21  3.6  1010 s-1 t Thus, 3.6  1010 nuclei decay in one second.

Now,

Natural radioactive processes are of two kinds: (1)   decay associated with the emission of   particles, viz. nuclei 42 He of helium. Alpha particles are heavy positively charged particles having a mass m  4 amu and a charge q = +2e. The velocity of -particles is relatively low: v = (c/30 - c /15), where c is the velocity of light. (2) decay (beta-minus-decay) associated with the emission of electrons formed at the instant of decay.  Both processes are accompanied by radiation, i.e. the flow of photons  having a very small wavelength, and hence a very high energy. Like other electromagnetic waves, -rays propagate  at a velocity of light. The penetrability of -rays is 0-100 times higher than the penetrability of -rays and 1000-10000 Lead times higher than the penetrability of -rays. It also exceeds the penetrability of X-rays. Fig.(12) In a magnetic field,  rays are In a magnetic field, a beam of -, -, undeviated and  - particles are the most deviated. and -rays splits into three parts. Nuclei possessing the artificial radioactivity are obtained by bombarding stable nuclei of heavy elements by -particles, neutrons, or (sometimes) protons and other particles. Nuclear transformations occur in two stages in this case. First a particle hits a target nucleus and causes its transformation into another, unstable (radioactive), nucleus. This newly formed nucleus spontaneously emits a particle and is transformed either into a stable nucleus or into a new radioactive nucleus. Artificial radioactivity obeys the same laws as natural radioactivity. Radioactive processes occur in accordance with the laws of conservation of energy, momentum, angular conservation, electric charge, and mass number (amount of nucleons). In -decay, the mass number of the nucleus decreases by four and the charge decreases by two units, as a result of which two electrons are removed from the atomic shell. The element transforms into another element with the atomic number which is two units lower. In --decay, a neutron in the nucleus transforms into a proton. Such a transformation of the neutral neutron into the positive proton is accompanied by the birth of an electron, i.e. by -radiation. The mass number of the nucleus does not change in this process, while the charge increases by +e and atomic number increases by one.



5.

ATOMIC NUCLEUS

The atomic nucleus consists of two types of elementary particles, viz. Protons and neutrons. These particles are called nucleons. The proton (denoted by p) has a charge +e and a mass mp  1.6726  10-27 kg, which is approximately 1840 times larger than the electron mass. The proton is the nucleus of the simplest atom with Z = 1, viz. the hydrogen atom. The neutron (denoted by n) is an electrically neutral particle (its charge is zero). The neutron mass mn  1.6749  10-27 kg. The fact that the neutron mass exceeds the proton mass by about 2.5 electron masses is of essential importance. It follows from this that the neutron in free state (outside the nucleus) is unstable (radioactive). During the time equal on the average to 12 min, the neutron spontaneously transforms to the proton by emitting an electron (e-) and a particle called the antineutrino ( v~ ). This process can be schematically written as follows: n  p + e- + v~ . The most important characteristics of the nucleus are the charge number Z (coinciding with the atomic number of the element) and the mass number A. The charge number Z is equal to the number of protons in the nucleus, and hence it determines the nuclear charge equal to Ze. The mass number A is equal to the number of nucleons in the nucleus (i.e. to the total number of protons and neutrons). Nuclei are symbolically designated as X ZA or ZXA where X stands for the symbol of a chemical element. For example, the nucleus of the oxygen 18 atom is symbolically written as O18 8 or 8O . Most of the chemical element have several types of atoms differing in the number of neutrons in their nuclei. These varieties are called isotopes. For example, oxygen has three stable isotopes: 17 18 O16 8 , O 8 and O8 . In addition to stable isotopes, there also exist unstable (radioactive) isotopes. Atomic masses are specified in terms of the atomic mass unit or unified mass unit (u). The mass of a neutral atom of the carbon isotope 6C12 is defined to be exactly 12 u. 1u = 1.66056  10-27 kg = 931.5 MeV Example 14 (a) Calculate the value of 1 u from Avogadro’s number. (b) Determine the energy equivalent of 1u. Solution (a) One mole of C12 has a mass of 12 g and contains Avogadro’s number, NA, of atoms. By definition, each C12 has a mass of 12 u. Thus, 12 g corresponds to 12 NA u which means 1g 1 1u =  N A 6.022045  10 23 or 1u = 1.66056  10-27 kg

 (b) From Einstein relation E = mc2  E = (1.66056  10-27) (3  108)2 = 1.4924  10-10 J Since 1eV = 1.6  10-19 J  E = 931.5 MeV Hence 1u = 931.5 MeV The shape of nucleus is approximately spherical and its radius is approximately related to the mass number by R  1.2 A1/3 fm where 1 fermi (fm) = 10-15 m Example 15  Find the mass density of the oxygen nucleus 8O16. Solution 4 4 Volume V = R3   1.2 3 A = 1.16  10-43 m3 3 3 Mass of oxygen atoms (A = 16) is approximately 16 u. m Therefore, density is  = v 27 161.66 10  = 2.3  1017 kg/m3 or = 1.16  10  43 14 This is 10 times the density of water.

Binding Energy The rest mass of the nucleus is smaller than the sum of the rest masses of nucleons constituting it. This is due to the fact that when nucleons combine to form a nucleus, the binding energy of nucleons is liberated. The binding energy is equal to the work that must be done to split the nucleus into particles constituting it. The difference between the total mass of the nucleons and the mass of the nucleus is called the mass defect of the nucleus: m = [Zmp + (A  Z)mn]  mnuc. Multiplying the mass defect by the square of the velocity of light, we can find the binding energy of the nucleus: BE = mc2 = [(Zmp + (A – Z)mn) – mnuc]c2 J (25) If the masses are taken in atomic mass unit, the binding energy is given by BE = [ZmP + (A – Z)mn – mnuc]931.5 MeV (26) Dividing the binding energy by the number A of nucleons in the nucleus, we obtain the binding energy per nucleon. Fig.(13) shows the dependence of the binding energy per nucleon BE/A on the mass number A of the nucleus. Nucleons in nuclei with mass numbers from 50 to 60 have the highest binding energy. The binding energy per nucleon for these nuclei amounts to 8.7 MeV and gradually decreases with increasing A. For the heaviest natural element  uranium  it amounts to 7.5 MeV. Figure shows that when a heavy nucleus (with A  240) splits into two nuclei with A = 120, the released energy is of

 the order of 1 MeV per nucleon i.e. 240 MeV per parent nucleus. It should be mentioned for comparison that when a carbon atom is oxidized (burnt) to CO2, the energy of the order of 5 eV is liberated, which is smaller than the energy released in fission of a uranium nucleus by a factor of 50 millions. BE 9 nucleon

56

Fe

238

U

7 (MeV)

5 3 1 80 120 160 200 240 A Fig.(13) The average binding energy per nucleon as a function of atomic number A. The maximum value occurs at Fe56. 40

It also follows from Fig(13) that the fusion (synthesis) of light nuclei into one should be accompanied by the liberation of a huge energy. For example, the fusion of two nuclei of heavy hydrogen 1 H 2 (this nucleus is called a deuteron) into a helium nucleus 2 He 4 would yield an energy equal to 24 MeV. The forces binding nucleons in a nucleus manifest themselves at distances < 10-15 m. In order to bring together two positively charged deuterons to such a distance, their Coulomb repulsion should be overcome. For this, the deuterons must have a kinetic energy equivalent to their mean energy of thermal motion at a temperature of the order of 109 K. For this reason, the fusion reaction of nuclei is also called a thermonuclear reaction. Actually, some thermonuclear reactions may occur at a temperature of the order of 107K. This is due to the fact that there is always a certain number of nuclei whose energy considerably exceeds the mean value. Example 16 Find the binding energy of 12 6 C ? Also find the binding energy per nucleon. Solution One atom of 12 6 C consists of 6 protons, 6 electrons and 6 neutrons. The mass of the uncombined protons and electrons is the same as that of very small binding energy of each proton-electron pair). Mass of six 11 H atoms = 6  1.0078 = Mass of six neutrons = 6  1.0087 = Total mass of component particles = Mass of 612C atom = Mass defect = Binding energy = (931)(0.099) = Binding energy per nucleon

=

six 11 H atoms (if we ignore the 6.0468 u 6.0522 u 12.0990 u 12.00004 0.0990 u 92 MeV 92  7.66 MeV 12

 Example 17 A neutron breaks into a proton and electron. Calculate the energy produced in this reaction in MeV. Mass of an electron = 9  10-31kg, Mass of proton = 1.6725  10-27kg, Mass of neutron 1.6747  10-27kg. Speed of light = 3  108 m/s. Solution 1 1 o on  1H + -1e Mass defect (m) = [Mass of neutron – (mass of proton + mass of electron)] = [1.6747  10-27 – (1.6725  10-27 + 9  10-31)] = 0.0013  10-27 kg  Energy released Q = m c2 Q = (0.0013  10-27)  (3  108)2 = 1.17  10-13 J 1.17  10 13 = = 0.73  106eV = 0.73 MeV 1.6  10 19

NUCLEAR REACTION A nuclear reaction in which a collision between particle a and nucleus X produces Y and particle b is represented as a + X  Y + b The reaction is sometimes expressed in the shorthand notation X(a, b)Y. Reactions are subjected to the restrictions imposed by the conservation of charge, energy, momentum and angular momentum. Energy of A Reaction a m1

Y

X K1

m2

Initial

K2

m3

b K3 Final

m4

K4

Initial energy: Ei = m1c2 + m2c2 + K1 + K2 Final energy: Ef = m3c2 + m4c2 + K3 + K4 Since Ei = Ef (energy conservation)  [(m1 + m2) - (m3 + m4)]c2 = (K3 + K4) - (K1 + K2) The energy, that is released or absorbed in a nuclear reaction is called the Q - value or disintegration energy of the reaction. Q = [(m1 + m2) - (m3 + m4)]c2 J (27 a) or Q = [(m1 + m2) - m3 + m4)] 931.5 u (27 b) If Q is positive, rest mass energy is converted to kinetic mass energy, radiation mass energy or both, and the reaction is exoergic. If Q is negative, the reaction is endoergic. The minimum amount of energy that a bombarding particle must have in order to initiate an endoergic reaction, is called Threshold Energy Eth . m  Eth = -Q  1  1 (28)  m2  where m1 = mass of the projectile m2 = mass of the target.

 Example 18 Neon - 23 beta decays in the following way : 23 10 Ne

23 11 Na  o1 e  

Find the minimum and maximum kinetic energy that the beta particle o1 e can have. The atomic masses of 23Ne and 23Na are 22.9945 u and 22.9898 u, respectively.

Solution 23 10 Ne

Reactant 22.9945 - 10me

Products 23 11 Na 22.9898 - 11 me o 1

22.9945 – 10 me

Total

e

-me

Total 22.9898 – 10 me

Mass defect = 22.9945 - 22.9898 = 0.0047 u Q = (0.0047)(931) = 4.4 MeV The  - particle and neutrino share this energy. Hence the energy of the -particle can range from 0 to 4.4 MeV. Example 19 How much energy must a bombarding proton possess to cause the reaction. 7 3 Li

11 H  74 Be  10 n

Solution Since the mass of an atom include the masses of the atomic electrons, the appropriate number of electron masses must be subtracted from the given values. Reactants Products 7 7 7.01600 - 3 me 7.01693 - 4me 3 Li 4 Be 1 1H

1.0783 - 1 me

1 on

1.0866

Total 8.02383 - 4me Total 8.02559 - 4me The Q-value of the reaction Q = -0.00176 u = 1.65 MeV The energy is supplied as kinetic energy of the bombarding proton. The incident proton must have more than this energy because the system must possess some kinetic energy even after the reaction, so that momentum is conserved. With momentum conservation taken into account, the minimum kinetic energy that the incident particle can be found with the formula. m   1 Eth = - 1  Q  1   1.65  1.89 MeV  M  7 Example 20 In a nuclear reactor, fission is produced in 1 g for U235(235.0439 u) in 24 hours by a slow neutron (1.0087 u). Assume that 35Kr92 (91.8973 u) and 56Ba141 (140.9139 amu) are produced in all reactions and no energy is lost. (a) Write the complete reaction (b) Calculate the total energy produced in kilowatt hour. Given 1 u = 931 MeV.

 Solution The nuclear fission reaction is 92U235 + on1  56Ba141 + 36Kr92 + 3 on1 Mass defect m = [(mu + mn) – (mBa + mKr + 3mn)] m = 256.0526 – 235.8373 = 0.2153 u Energy released = 0.2153  931 = 200 MeV 6.02  10 23 Number of atoms in 1 g = = 2.56  1021 235 Energy released in fission of 1 g of U235 is Q = 200  2.56  1021 = 5.12  1023 MeV = (5.12  1023)  (1.6  10-13) = 8.2  1010 J 8.2  1010 = kWh = 2.28  104 kWh 6 3.6  10 Example 21 It is proposed to use the nuclear fusion reaction: 2 2 4 1H + 1H = 2He in a nuclear reactor of 200 MW rating. If the energy from above reaction is used with a 25% efficiency in the reactor, how many grams of deuterium will be needed per day. (The masses of 1H2 and 2He4 are 2.0141 and 4.0026 u respectively). Solution Energy released in the nuclear fusion is Q = mc2 = m(931)MeV  Q = (2  2.0141 – 4.0026)  931MeV = 23.834 MeV = 23.834  106eV Since efficiency of reactor is 25% So effective energy used = 25 / 100  23.834  106  1.6  10-19 J = 9.534  10-13 J Since the two deuterium nucleus are involved in a fusion reaction, therefore, energy 9.534  10 13 released per deuterium is 2 For 200MWpower per day 200  10 6  86400 number of deuterium nuclei required = = 3.624  1025 9.534  10 13 2 Since 2g of deuterium constitute 6  1023 nuclei, therefore amount of deuterium required is 2  3.624  10 25 g= = 120.83 g/day 6  10 23

6.

PHOTOELECTRIC EFFECT

The emission of electrons from a metallic surface when irradiated by electromagnetic radiation is called the phenomenon of photoelectric effect. The emitted electrons are called as photoelectrons.

 L For the investigation of the photoelectric ` effect a schematic diagram of the apparatus as used by Lenord (1902) is shown in the fig.(14). Monochromatic light from the lamp L illuminates a plate P in an evacuated glass enclosure. A C P e battery maintains a potential difference between P and a metal cylinder C, which collects the photoelectrons. The potential C can be varied to be either positive or negative relative to P. When the collector is positive with respect to the  + A plate, the electrons are attracted to it and V the ammeter (A) registers a current. Fig.(14) Schematic diagram of apparatus used for the investigation of the photoelectric effect. Lenard studied the dependence of L - lamp; P – metal plate, photoelectric current on the following C – conducting cylinder factors. (i) Intensity of incident radiation (ii) Potential difference between the plate and the collector (iii) Frequency of the incident radiation The result of observations are as follows:

Effect of the intensity of Incident Radiation When the collector is positive relative to the plate and the potential difference is kept fixed, then for a given frequency of radiation, the photoelectric current is proportional to the intensity of the light, as shown in fig.(15). It shows that the number of emitted photoelectrons is proportional to the light intensity. Furthermore, there is no threshold intensity.

i (mA)

O

Intensity (I) Wm-2 Fig.(15)

Effect of Potential Difference When the frequency and intensity of radiation are kept constant and the positive potential of collector relative to plate is gradually increased, then the photoelectric current i increases with the potential difference V. At some value of the potential difference, when all the emitted electrons are collected, thus increasing potential difference has no effect on the current. The current has reached its maximum value, called the saturation current. When the polarity of the battery is reversed, the electrons are repelled and only the most energetic ones reach the collector, so the current falls. When the retarding potential difference reaches a critical value, the current drops to zero. At this stopping potential Vo, only those electrons with the maximum kinetic energy are able to reach the collector. 1 2 eVo = mv max (29) 2

 For a given frequency of light, the saturation current depends on the intensity of light. Larger the intensity; higher the saturation current. However, the stopping potential does not change with the intensity. It is clearly shown in fig. (16).

Effect of frequency For a given intensity of radiation, the stopping potential depends on the frequency. Higher the frequency, higher the value of stopping potential.

i I2 I1 I2 > I1

O V Vo Fig.(16) At positive accelerating potential differences, the maximum current is determined by the intensity of the radiation. However, the stopping potential does not change with the intensity.

The maximum kinetic energy of the electrons depends on the light source and the plate material, but not on the intensity of the source. Certain combinations of light sources and plate materials exhibit no photoelectric effect. 1 2 mv max 2

Saturation Current

(1)

(2)

f2 f1 Vo2 Vo1 O V Fig.(17) For a given intensity, stopping potential depends on frequency. If f2 > f1, Vo2 > Vo1.

O

Fig.(18)

fo1

fo2

V The maximum kinetic energy is proportional to the frequency of light. fo1 - threshold frequency of metal (1) fo2 – threshold frequency of metal (2)

For a metal plate there exists a minimum frequency called threshold frequency (fo) below which no electron is emitted however large the light intensity may be. The threshold frequency is a characteristic of the metal plate.

Einstein’s Theory of Photoelectric Effect According to Einstein, the experimental results of photoelectric effect can be explained by applying the quantum theory of light. He assumed that light of frequency f contain packets or quanta of energy E = hf. On this basis, light consists of particles, and these are called photons. The number of photons per unit area of cross-section of the beam of light per second is proportional to its intensity. But the energy of photon is proportional to its frequency and is independent of the light intensity.

 In the process of photo emission a single photon gives up all its energy to a single electron. As a result, the electron is ejected instantaneously. Since the intensity of light is determined by the number of photons incident, therefore, increasing the intensity will increase the number of ejected electrons. 1 2  The maximum possible kinetic energy  mv max  of the photoelectrons is determined by 2  the energy of each photon (hf) according to the Einstein equation (30) 1 2 mv max  hf  W (30) 2 where the work function, (W), is the minimum energy needed to extract an electron from the surface of the material. In terms of threshold frequency, it is given by W = hfo (31) Using equation (31), we may write equation (30) as 1 2 (32) mv max = hf – hfo = h(f – fo) 2 Also, in terms of stopping potential, eVo = h(f – fo) (33) Example 22 Ultraviolet light of wavelength 2000Å causes photoemission from a surface. The stopping potential is 2V. (a) Find the work function in eV (b) Find the maximum speed of the photoelectrons. Solution (a) Using Einstein relation hc W=  eVo  12400 or W=  2  4.2eV 2000 1 2 (b) Since mv max  eVo 2  or





2eVo 2 1.6  10 19 2    m 9.1  10 31 vmax = 8.4  105 m/s vmax =

How to determine the photoelectric current? Let P be the power of a point source of electromagnetic radiations, then intensity I at a distance r from the source is given by P I= (W/m2) (34) 2 4r

 If A is the area of a metal surface on which radiations are incident, then the power received by the plate is  P  P = IA =  (35)  A (W)  4r 2  If f is the frequency of radiation, then the energy of photon is given by E = hf

Area (A)

r Fig. 19

The number of photons incident on the plate per second (called photon flux) is given by

 P   A  2 P'  =   4r E  hf   

(36)

If f > fo (threshold frequency) and photon efficiency of the metal plate is %, then the number of photoelectrons emitted per second is given by  P   A  2    n= (37)   4r 100  hf  100   Finally, the photocurrent i is given by i = ne (38) where e is the charge of an electron (e = 1.6  10-19 J) Example 23 The intensity of sunlight on the surface of earth is 1400 W/m2. Assuming the mean wavelength of sunlight to be 6000Å, calculate (a) the photon flux arriving at 1 m2 area on earth perpendicular to light radiations, and (b) the number of photons emitted from the sun per second assuming the average radius of Earth’s orbit is 1.49  1011 m. Solution hc 12400 (a) Energy of a photon E =   2.06 eV  3.3  0 19 J  6000  14001 IA  Photon flux = = 4.22  1021 19 E 3.3  10 P I 4R 2 (b) n=  E E 140041.49 1011 2 = 1.18  1045 or n= 3.3  10 19







Objective Solved Examples 1.

The intensity of X-rays from a Coolidge tube is plotted against wavelength  as shown in the figure. The minimum wavelength found is C and the wavelength of the K line is K. As the accelerating voltage is increased (a) K - C increases (b) K - C decreases (c) K increases (d) K increases 

I

C

K



Solution As the accelerating voltage is increased, C decreases while K remains the same.  (a) The half-life of 215At is 100 s. The time taken for the radioactivity of a sample of 215At to decay to 1/16th of its initial value is (a) 400 s (b) 6.3 s (c) 40 s (d) 300 s Solution For the decay of 1/16th of initial value, four half lives are required.  (a) 2.

3.

Which of the following processes represents a gamma-decay? (a) A X Z    A X Z 1  a  b (b) A X Z  1 n o  A3 X Z  2  c

(c) A X Z  A X Z  f (d) A X Z  e 1  A X Z 1  g Solution In gamma-decay, the atomic and mass number do not change.  (c) r , Vo and ro are constants and ro r is the radius of orbit. The radius rn of the nth Bohr's orbit depends on principal quantum number n as 1 (a) rn  n (b) rn  2 n 1 (c) rn  n 2 (d) rn  n Solution r V  Vo log e ro dV  Vo F   dr r

4.

The attractive potential for an atom is given by V  Vo ln



mV 2 Vo  r r 2 mv  Vo nh mvr  2 m 2V 2 r 2 n 2 h 2 / 4  2  Vo mV 2



r 2  n2 r  n (a) 

5.

Volume V of a nucleus is related to the mass M as (a) V  M3 (b) V  3 M 1 (c) V  M (d) V  M Solution Radius of nucleus is given by r  A1 / 3 ro 4 volume = r 3  v  A  M 3  (c)  The ratio of radius of 100Fm257 atom (assuming Bohr's model to be valid) to the Bohr radius is (a) 257 (b) 100 1 (c) (d) 4 4 Solution Since the outermost orbit number is 5 and n2 1 rn = rH = rH Z 4  (c) 6.

A stationary nucleus (mass number 220) decays by emitting an -particle. The total energy released is 5.5 MeV. The kinetic energy carried by the -particle is (a) 5.4 MeV (b) 5.6 MeV (c) 4.9 MeV (d) 6.5 MeV Solution Applying law of conservation of linear momentum and using the energy released, velocity of -particle and daughter nucleus can be calculated. Therefore, kinetic energy of  particle 220  4 =  5.5 MeV = 5.4 MeV. 220  (a) 7.

 8.

After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is (a) 6000 (b) 9000 (c) 3000 (d) 24000 Solution The half-life of the substance is 140 days. In 420 days, there will be three half-lives.  (d) 9.

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let 1 be the de-Broglie wavelength of the proton and 2 be the wavelength of the photon. The ratio 1/2 is proportional to (a) Eo (b) E1/2 (c) E-1 (d) E-2 Solution h hc 1 = and 2  E 2mE  (b) 10.

The figure shows the variation of Photo current photocurrent with anode potential for a photo-sensitive surface for three different radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc be the frequencies for the c curves a, b and c respectively. b (a) fa = fb and Ia  Ib (b) fa = fc and Ia = Ic (c) fa = fb and Ia = Ib O (d) fb = fc and Ib = Ic  Solution The stopping potential for curves a and b is same.  fa = fb Also saturation current is proportional to intensity.  Ia < Ib  (a) 11.

a

Anode potential

A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector? (a) 2 photon of energy 10.2 eV (b) 2 photon of energy of 1.4 eV (c) One photon of energy 10.2 eV and an electron of energy 1.4 eV (d) One photon of energy 10.2 eV and another photon of 1.4 eV

 Solution Due to 10.2 eV photon one photon of energy 10.2 eV will be detected. Due to 15 eV photon the electron will come out of the atom with energy (15 - 13.6) = 1.4 eV.  (c) 12.

K wavelength emitted by an atom  is given by an atom of atomic number Z = 11 is . Find the atomic number for an atom that emits K radiation with wavelength 4  (a) Z = 6 (b) Z = 4 (c) Z = 11 (d) Z = 44

Solution f1 

v  a 11  1 1

f2 

v  aZ  1 2

By dividing, 2 10  1 Z  1    13.

1 10  4 Z 1 Z=6 (a)

If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is [Mass of He nucleus is 4.0026 amu and mass of Oxygen nucleus is 15.9994 amu] (a) 7.6 MeV (b) 56.12 MeV (c) 10.24 MeV (d) 23.9 MeV Solution Mass defect = 4  4.0026 - 15.9994 = 0.011 Energy released by oxygen nuclei = 0.011  931 = 10.24 MeV  (c)