Chapter 2 Wave Nature of Matter (Pp 43-72)

CH 02

WAVE NATURE OF MATTER

43

CHAPTER 2 WAVE NATURE OF MATTER 2-1 DE BROGLIE WAVELENGTH Problem 2-1 A bullet of mass 41 g travels at 960 m s-1. (a) What wavelength can we associate with it? (b) Why does the wave nature of the bullet not reveal itself through the diffraction effects? Solution (a) The de Broglie wavelength is given by h (6.626 × 10 −34 ) λ= = = 1.683 × 10 −35 m −3 mv (41 × 10 )(960) (b) This value of wavelength is extremely small as compared with any reasonable microscopic distance, therefore the wave nature of the bullet is not revealed through diffraction effects. Problem 2-2 What will be the de Broglie wavelength associated with a mass of 0.01 kg moving with a velocity of 10 m/s? K.U. B.Sc. 2002 Solution The de Broglie wavelength associated with a moving object is given by h 6.626 × 10 −34 λ= = = 6.626 × 10 −33 m mv (0.01)(10) Problem 2-3 Calculate the de Broglie wavelength of virus particle of mass 1 × 10 −15 kg moving at a speed of 2 mm/s. P.U. B.Sc. 2005

CH 02 WAVE NATURE OF MATTER 44 Solution The de Broglie wavelength is given by h (6.626 × 10 −34 ) λ= = = 3.313 × 10 −16 m −15 −3 mv (1 × 10 )(2 × 10 ) Problem 2-4 A tennis ball having mass 20 gm is moving at a speed of 50 m/s. Compare its de Broglie wavelength to that of an electron moving with a velocity of 5 × 10 6 m/s. Solution The de Broglie wavelength associated with a moving object is h given by λ= mv For tennis ball h (6.626 × 10−34 ) λ= = = 6.626 × 10− 34 m −3 m v (20 × 10 )(50) This wavelength is too small to be observed experimentally. For de Broglie wavelength of electron we have

(6.626 × 10 −34 ) h λe = = = 1.455 × 10 −10 m − 31 6 m0 v (9.109 × 10 )(5 × 10 ) Now

λ 6.626 × 10−34 = = 4.554 × 10− 24 −10 λe 1.455 × 10

Problem 2-5 Find the de Broglie wavelength of 1.0 mg grain of sand blown by the wind at a speed of 20 m/s. Solution The de Broglie wavelength is given by h 6.626 × 10−34 λ= = = 3.313 × 10− 29 m −6 m v (1.0 × 10 )(20)

CH 02 WAVE NATURE OF MATTER 45 Problem 2-6 Calculate the de Broglie wavelength of an electron moving with a speed of 9 x 106 m s -1. B.U. B.Sc. 2002A Solution The de Broglie wavelength is given by h (6.626 × 10 −34 ) λ= = = 8.082 × 10 −11 m − 31 6 m0 v (9.109 × 10 )(9 × 10 ) = 0.808 Ǻ Problem 2-7 The speed of electron in first orbit of Hydrogen atom is 2 × 106 m/s. Calculate the de Broglie wavelength. Solution The de Broglie wavelength is given by h (6.626 × 10−34 ) = λ= m0v (9.109 × 10− 31 )(2 × 106 )

λ = 3.64 × 10 −10 m = 3.64 Ǻ This wavelength is comparable to the size of atom and should be observable. Problem 2-8 If the de Broglie wavelength of an electron is 1.1 × 10 −10 m, what is the speed of the electron? P.U. B.Sc. 2006 Solution The de Broglie wavelength is given by h λ= m0 v h (6.626 × 10 −34 ) v= = = 6.613 × 106 m / s −31 −10 m0 λ (9.109 × 10 )(1.1×10 ) Problem 2-9 What will be the speed of an electron with an associated wavelength of 2.00 Å? Solution The de Broglie wavelength is defined as

CH 02

WAVE NATURE OF MATTER h h λ= = p m0v h v= m0 λ

46

(6.626 × 10 −34 ) v= = 3.64 × 106 m / s − 31 −10 (9.109 × 10 )(2.00 × 10 ) Problem 2-10 Calculate the de Broglie wavelength of a 1.00 keV (a) electron, (b) neutron and (c) a photon. Solution 1 2 Now mv = K 2 mv 2 = 2 K m 2 v 2 = 2mK or p 2 = 2mK or p = 2mK h h λ= = p 2mK (a) The de Broglie wavelength for an electron is given by h 6.626 × 10 −34 = λ= 2me K 2(9.109 × 10 −31 )(1 × 10 3 × 1.602 × 10 −19 ) = 3.88 × 10 −11 m or 38.8 pm (b) For a neutron m is replaced by mn i.e. h 6.626 × 10 −34 λ= = 2m n K 2(1.675 × 10 − 27 )(1 × 10 3 × 1.602 × 10 −19 ) = 9.04 × 10 −13 m or 904 fm (a) The energy of a photon is given by relation hc E=

λ

or

λ=

hc (6.626 × 10 −34 )(2.998 × 10 8 ) = E (1 × 10 3 × 1.602 × 10 −19 )

λ = 1.24 × 10 −9 m

or 1.24 nm

CH 02 WAVE NATURE OF MATTER 47 Problem 2-11 Calculate the wavelength associated with an alpha particle emitted by the nucleus of an atom of Radon-226. The kinetic energy of alpha particle emitted from Radon-222 is 5.486 MeV. Solution The de Broglie wavelength corresponding to this energy is h λ= 2mα K

λ=

6.626 × 10 −34 2(6.445 × 10 − 27 )(5.486 × 10 6 × 1.602 × 10 −19 )

Q 1 eV = 1.602 × 10 −19 J λ = 6.225 × 10 −15 m

Problem 2-12 Find the de Broglie wavelength of a 40 keV electrons used in a certain electron microscope. Solution The de Broglie wavelength corresponding to kinetic energy ‘K’ is h 6.626 × 10 −34 λ= = 2m0 K 2(9.109 × 10 −31 )(50 × 10 3 × 1.602 × 10 −19 )

λ = 6.133 ×10 −12 m Problem 2-13 Find the wavelength of an electron which has been accelerated by a potential difference of 54 volts. B.U. B.Sc. 1989A, 1992A Solution 1 Now K .E. = m0v 2 = V0e 2 2 m0v = 2V0e or m02v 2 = 2m0V0e

CH 02

WAVE NATURE OF MATTER 48 p = m0v = 2m0V0e The de Broglie wavelength is given by (6.626 × 10−34 ) h h λ= = = p 2m0V0e 2(9.109 × 10−19 )(54)(1.602 × 10−19 )

λ = 1.669 × 10 −10 m

or 1.669 Ǻ

Problem 2-14 Suppose we are to build an electron microscope and we want to operate at a wavelength of 0.10 nm. What accelerating should be used? Solution 1 Now V0 e = m 0 v 2 2 m v2 m v2 p2 (h / λ ) 2 V0 = 0 = 0 = = 2 e 2 m0 e 2 m 0 e 2 m 0 e

V0 =

h2 2 m0e λ2

(6.626 × 10 −34 ) 2 = 150 volts 2(9.109 × 10 −31 )(1.602 × 10 −19 )(0.10 × 10 −9 ) 2 Problem 2-15 Green light has a wavelength of about 550 nm. Through what potential difference must an electron be accelerated to have this wavelength? Solution 1 V0 e = m 0 v 2 Now 2 m v2 m v2 p2 (h / λ ) 2 V0 = 0 = 0 = = 2 e 2 m0 e 2 m 0 e 2 m 0 e V0 =

V0 =

h2 2 m0e λ2

CH 02

WAVE NATURE OF MATTER (6.626 × 10 −34 ) 2 V0 = 2(9.109 × 10 −31 )(1.602 × 10 −19 )(550 × 10 −9 ) 2

49

V0 = 4.97 × 10 −6 volts = 4.97 µ V Problem 2-16 Calculate the de Broglie wavelength of 5 MeV alpha particles. P.U. B.Sc. 1990 Solution As given the energy of alpha particle, 5 MeV, is much smaller than its rest mass energy i.e. mα c 2 = 3727 MeV , therefore it is a non-relativistic case. The de Broglie wavelength is given by h h λ= = p 2mα K

where h = 6.626 × 10 −34 Js ,

mα = 6.645 × 10 −27 kg and

K = 5MeV = 5 × 10 6 eV = (5 × 10 6 )(1,602 × 10 −19 ) J = 8.01 × 10 −13 J Hence 6.626 × 10 −34 λ= = 6.836 × 10 −15 m = 6.836 fm − 27 −13 2(6.645 × 10 )(8.01 × 10 ) Problem 2-17 A proton and electron have the same de Broglie wavelength. How does their speeds compare, assuming both are much less than that of light? Solution Now λ p = λe h h = m p v p me v e ve m p 1.673 × 10 −27 = = = 1.836 × 10 3 −31 v p me 9.109 × 10 The velocity of electron is 1836 times the velocity of proton.

CH 02 WAVE NATURE OF MATTER 50 Problem 2-18 Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the speeds are nonrelativistic. Solution The kinetic energy of a particle is given by 1 m2v 2 p2 (h / λ ) 2 h2 K = m v2 = = = = 2 2 m 2 m 2 m 2 m λ2 Now h2 Ke = 2 me λ 2

h2 Kp = 2 m p λ2 Therefore

Q the wavelengths are equal.

K e ( h 2 / 2 me λ2 ) m p = = K p ( h 2 / 2 m p λ2 ) me

K e 1.673 × 10 −27 = = 1837 K p 9.109 × 10 −31 Problem 2-19 Find the kinetic energy of an electron whose de Broglie wavelength is the same as that of 100 keV X-ray. B.U. B.Sc. 2009S Solution The energy of the given X-ray photon is E = 100 keV = 100 × 10 3 eV = (1 × 10 5 )(1.602 × 10 −19 ) J E = 1.602 × 10 −14 J h c E= But

λ

h c (6.626 × 10 −34 )(2.998 × 10 8 ) = = 1.239 × 10 −11 m −14 E 1.602 × 10 This is the de Broglie wavelength of given electron. The kinetic energy of the electron is given by

λ=

CH 02

WAVE NATURE OF MATTER m02 v 2 1 p2 (h / λ ) 2 h2 2 K = m0 v = = = = 2 2 m 0 2 m0 2 m0 2 m0 λ 2

51

(6.626 × 10 −34 ) 2 K= = 1.570 × 10 −15 J − 31 −11 2 2(9.109 × 10 )(1.239 × 10 ) 1.570 × 10 −15 K= eV = 9.80 × 10 3 eV = 9.80 keV −19 1.602 × 10 Problem 2-20 Calculate the kinetic energy of a proton having a de Broglie wavelength of 0.4 Å. Solution The kinetic energy of the proton is given by m 2p v 2 1 p2 2 K = mpv = = 2 2 mp 2 mp K=

(h / λ ) 2 h2 = 2 mp 2 m p λ2

Q p=

h

λ

−34 2

K= K=

(6.626 × 10 ) = 8.201 × 10− 20 J − 27 −10 2 2(1.673 × 10 )(0.4 × 10 )

8.201 × 10−20 eV = 0.512 eV 1.602 × 10−19

Problem 2-21 Calculate the de Broglie wavelength of an electron whose kinetic energy is 120 eV. P.U. B.Sc. 2000, 2003, F.P.S.C. 2007, K.U. B.Sc. 2007 Solution The de Broglie wavelength is given by h h λ= = p 2m 0 K

λ=

6.626 × 10 −34 2(9.109 × 10 −31 )(120 × 1.602 × 10 −19 )

CH 02

WAVE NATURE OF MATTER 52 −19 Q 1 eV = 1.602 × 10 J −10 λ = 1.12 × 10 m or 1.12 Ǻ

Problem 2-22 Calculate the de Broglie wavelength of electrons with kinetic energy 500 eV. Solution The de Broglie wavelength of electrons with kinetic energy ‘K’ is given by h h λ= = p 2m0 K

λ=

6.626 × 10−34 2(9.109 × 10− 31 )(500 × 1.602 × 10−19 )

λ = 5.485 ×10 −11 m = 0.055 nm Problem 2-23 Calculate the de Broglie wavelength of an electron whose kinetic energy is 1 keV. P.U. B.Sc. 2007 Solution 1 Now m0 v 2 = K 2 m0 v 2 = 2 K m02 v 2 = 2m0 K p 2 = 2m 0 K

λ= λ=

or

p = 2 m0 K

h h = p 2m0 K 6.626 × 10 −34

2(9.109 × 10 −31 )(1 × 10 3 × 1.602 × 10 −19 )

λ = 3.88 ×10−11 m = 38.8 pm

CH 02 WAVE NATURE OF MATTER 53 Problem 2-24 Find the kinetic energy of an electron with the same wavelength as blue light (λ = 450 nm). Solution The kinetic energy of a particle is given by 1 m2v 2 p2 (h / λ ) 2 h2 2 K= m v = = = = 2 2 m 2 m 2 m 2 m λ2 Now h2 Ke = 2 me λ 2

Ke =

(6.626 × 10−34 ) 2 = 1.190 × 10− 24 J − 31 −9 2 2(9.109 × 10 )(450 × 10 )

1.190 × 10 −24 eV = 7.43 × 10− 6 eV 1.602 × 10 −19 Problem 2-25 Find the kinetic energy of a neutron with the same wavelength as blue light (λ = 450 nm). Solution The kinetic energy of a particle is given by 1 m2v 2 p2 (h / λ ) 2 h2 K = m v2 = = = = 2 2 m 2 m 2 m 2 m λ2 h2 Now K n = 2 mn λ2 Ke =

Ke =

(6.626 × 10−34 ) 2 = 6.472 × 10− 28 J − 27 −9 2 2(1.673 × 10 )(450 × 10 )

6.472 × 10−28 eV = 4.04 × 10− 9 eV −19 1.602 × 10 Problem 2-26 Find the de Broglie wavelength of electron, proton and alpha particle all having same kinetic energy of 100 eV. Ke =

CH 02 WAVE NATURE OF MATTER 54 Solution The de Broglie wavelength in terms of kinetic energy K is given by h h λ= = p 2mK

(6.626 × 10 −34

1.171× 10 −25 λ= = m 2m(100 × 1.602 ×10 −19 ) For electron m = me = 9.109 × 10 −31 kg and

λ=

1.171 × 10 −25

=

1.171 × 10 −25

9.109 × 10 −31 λ = 1.227 ×10 −10 m = 1.227 Ǻ For proton m = m p = 1.673 × 10 −27 kg and

λ=

m

1.171 × 10 −25

=

1.171 × 10 −25

1.673 × 10 −27 λ = 2.863 ×10 −12 m = 0.02863 Ǻ For electron m = mα = 6.695 × 10 −27 kg and

λ=

m

1.171 × 10 −25

=

1.171 × 10 −25

6.695 × 10 − 27 λ = 1.431×10 −12 m = 0.01431 Ǻ m

Problem 2-27 Calculate the de Broglie wavelength of an alpha particle accelerated through 600 volts. Solution Now T = V0e = (600)(1.602 × 10 −19 ) = 9.612 × 10 −17 J The de Broglie wavelength in terms of kinetic energy is given by

CH 02

h λ= = p

λ=

WAVE NATURE OF MATTER h

55

2mα K 6.626 × 10−34

2(6.645 × 10

− 27

)(9.612 × 10

−17

)

= 5.86 × 10−13 m

Problem 2-28 The wavelength of the yellow spectral emission line of sodium is 589 nm. At what kinetic energy would an electron have the same de Broglie wavelength? K.U. B.Sc. 2001, P.U. B.Sc. 2009 Solution The de Broglie wavelength is given by h h λ= = p 2m 0 K

Square both sides h2 λ2 = 2m0 K

or

K=

h2 2m0λ2

K=

(6.626 ×10 −34 ) 2 = 6.947 ×10 −25 J 2(9.109 ×10 −19 )(589 ×10 −9 ) 2

K=

6.947 × 10 −25 eV = 4.34 × 10 −6 eV −19 1.602 × 10

Problem 2-29 If the de Broglie wavelength of a proton is 0.113 pm, (a) What is the speed of the proton and (b) through what electric potential would the proton have to be accelerated from rest to acquire this speed? Solution (a) The de Broglie wavelength for proton is given by h h λ= or v = mpv mpλ

CH 02

WAVE NATURE OF MATTER 56 −34 (6.626 ×10 ) v= = 3.504 ×106 ms −1 − 27 −12 (1.673 ×10 )(0.113 ×10 ) (b) If Vo is the potential through which the proton has to be accelerated to acquire speed v, then 1 V0 e = m p v 2 2 m p v 2 (1.673 × 10 −27 )(3.504 × 10 6 ) 2 = or V0 = 2e 2(1.602 × 10 −19 )

V0 = 6.411 × 10 4 volts = 64.11kV Problem 2-30 A “thermal neutron” in a substance at temperature T is a neutron with kinetic energy equal to (3kT / 2) , where k = 1.381 × 10 −23 J / K is the Boltzmann constant. Determine the wavelength of a thermal neutron in a reactor at (a) 300 K and (b) 800 K. Solution The de Broglie wavelength is given by h h h h λ= = = = p 2mK 3mkT 3  2m kT  2  For a thermal neutron (6.626 × 10 −34 ) 2.515 × 10 −9 λ= = T 3(1.675 × 10 −27 )(1.381 × 10 − 23 )T

(a) T = 300 K , λ = (b) T = 800 K , λ =

2.515 × 10 −9 300 2.515 × 10 −9 800

= 1.452 × 10 −10 m = 8.893 × 10 −11 m

CH 02 WAVE NATURE OF MATTER 57 Problem 2-31 What is the de Broglie wavelength of an electron that is accelerated through a potential difference of 50 kV in a colour T.V. set? Solution The de Broglie wavelength is given by h h h λ= = = p 2mK 2mVe

λ=

6.626 × 10 −34 2(9.109 × 10 −31 )(50 × 103 )(1.602 × 10 −19 )

λ = 5.485 × 10 −12 m Problem 2-32 An electron, a proton and a lead atom have the same de Broglie wavelength λ = 280 pm. Determine the kinetic energy of each. Solution The de Broglie wavelength in terms of kinetic energy K is given by h h λ= = p 2mK Squaring both sides h2 λ2 = 2mK h2 (6.626 × 10−34 ) 2 2.800 × 10−48 = = or K = 2mλ2 2m(280 × 10−12 ) 2 m −31 For electron m = me = 9.109 × 10 kg and

2.800 × 10 −48 K= = 3.074 × 10 −18 J or 19.2 eV −31 9.109 × 10 For proton m = m p = 1.673 × 10 −27 kg and K=

2.800 × 10 −48 = 1.674 × 10 − 21 J − 27 1.673 × 10

or 19.2 eV

CH 02 WAVE NATURE OF MATTER For lead atom m = 3.454 × 10 −25 kg

K=

2.800 × 10 −48 = 8.107 × 10 − 24 J 3.454 × 10 −25

58

or 5.061 × 10 −5 eV

Problem 2-33 Calculate the kinetic energy of a neutron whose de Broglie wavelength is 10 -14 m. Solution The de Broglie wavelength in terms of kinetic energy K is given h h by λ= = p 2m n K

Squaring both sides λ2 = K=

h2 2m n K h2 2mn λ2

(6.626 × 10 −34 ) 2 = 1.311×10 −12 J − 27 −14 2 2(1.675 ×10 )(1× 10 ) K = 8.18MeV Q 1eV = 1.602 × 10 −19 J As rest mass energy of a neutron is 939 MeV, therefore the use of non-relativistic expression is justified. K=

Problem 2-34 Calculate the de Broglie wavelength of an electron having kinetic energy equal to 2 MeV. Solution Now K = 2 MeV = 2 × 10 6 eV K = (2 × 106 (1.602 ×10 −19 ) J = 3.204 × 10 −13 J The rest mass energy of the electron is m0 c 2 = (9.109 × 10 −31 )(2.998 × 10 8 ) 2 = 8.187 × 10 −14 J

CH 02 WAVE NATURE OF MATTER 59 The kinetic energy of electron is greater than its rest mass energy (about eight times), therefore the relativistic expression must be used i.e. E = K + m0 c 2

p 2 c 2 + m02 c 4 = K + m0 c 2 since E 2 = p 2 c 2 + m02 c 4 Square both sides p 2 c 2 + m02 c 4 = K 2 + 2 Km0 c 2 + m02 c 4 K 2 + 2 Km0 c 2 K ( K + 2m0 c 2 ) p = = c2 c2 2

p=

K ( K + 2m 0 c 2 ) c

Hence

λ=

h = p

λ=

hc K ( K + 2m0c 2 )

(6.626 × 10 −34 )(2.998 × 108 ) (3.204 × 10−13 ){( 3.204 × 10 −13 ) + 2(8.187 × 10 −14 )

λ = 5.044 × 10 −13 m Problem 2-35 An electron has a de Broglie wavelength 5.0 × 10 −10 m . (a) What is its momentum? (b) What is its speed? (c) What voltage was needed to accelerate it to this speed? B.U. B.Sc. 2009A Solution h 6.626 × 10 −34 (a) p = = = 1.325 × 10 − 24 kg • m / s −10 λ 5.0 × 10 h h (b) λ = = p m0 v

v=

h 6.626 × 10 −34 = = 1.455 × 106 m / s − 31 −10 m0 λ (9.109 × 10 )(5.0 × 10 )

CH 02 WAVE NATURE OF MATTER 60 6 v 1.455 × 10 (c) As = = 0.005