# Chapter 4 Atomic Structure (Pp 96-125)

January 15, 2018 | Author: Muhammad Ashfaq Ahmed | Category: Energy Level, Electron, Atomic Orbital, Atoms, Electromagnetic Spectrum

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CHAPTER 04

ATOMIC STRUCTURE

96

CHAPTER 4 ATOMIC STRUCTURE BOHR’S THEORY OF HYDROGEN ATOM Example 4-1 Determine the radius of the first orbit of the hydrogen atom. Solution The radius of electron in nth orbit of hydrogen atom is given by n2h2 rn = 4π 2 me 2 k (1) 2 (6.626 × 10 −34 ) 2 r1 = 4π 2 (9.109 × 10 −19 )(1.602 × 10 −19 ) 2 (8.988 × 10 9 ) r1 = 5.287 × 10 −11 m ≅ 0.53 Ǻ Example 4-2 Find the value of quantum number ‘n’ for a hydrogen atom that has an orbital radius of 847 pm. Solution The radius of in nth orbit of hydrogen atom is given by rn = 5.3 × 10 −11 n 2 in metres rn 847 × 10 −12 = = 4 the desired quantum 5.3 × 10 −11 5.3 × 10 −11 number. Example 4-3 Determine the speed of electron in 10th orbit of hydrogen atom. B.U. B.Sc. (Hons.) 1984A Solution The velocity of electron in nth orbit of hydrogen atom is given by 2π k e 2 vn = nh n=

CHAPTER 04 ATOMIC STRUCTURE 97 9 −19 2 2π (8.988 × 10 )(1.602 × 10 ) v10 = = 2.187 × 10 5 m / s − 34 (10)(6.626 × 10 ) the desired speed. Example 4-4 Find the de Broglie wavelength of an electron in the n = 3 orbit of hydrogen atom. In what region of the spectrum would a photon of the same wavelength be classified? Solution The velocity of electron in nth orbit of hydrogen atom is given 2π k e 2 2π k e 2 by vn = or v3 = for n = 3. 3h n h The de Broglie wavelength is defined as h h 3h 2 λ= = = p m v3 2π m k e 2

3(6.626 × 10 −34 ) 2 2π (9.109 × 10 −31 )(8.988 ×109 (1.602 × 10 −19 ) 2 λ = 9.977 × 10 −10 m ≅ 10 Å The photon of this wavelength belongs to X-ray region. Example 4-5 Calculate the frequency of electron in the first Bohr orbit of hydrogen atom. B.U. B.Sc. 2004A Solution The velocity of electron in terms of radius is given by nh v= 2π m r The orbital frequency is given by f = (speed of electron) ÷ (circumference of orbit) nh 1 nh f = × = 2π m r 2πr 4π 2 mr 2 n2h2 But r = therefore 4π 2 me 2 k

λ=

CHAPTER 04

ATOMIC STRUCTURE 2

98

 4π me k  nh  4π me k  =     3 3 4π 2 m  n 2 h 2   n h  2 −31 −19 4 4π (9.109 × 10 )(1.602 × 10 ) (8.988 × 10 9 ) 2 f = (1) 3 (6.626 × 10 −34 ) 3 2

2

2

4

2

f =

= 6.577 × 1015 Hz

Example 4-6 On the average, an atom will exist in an excited state for 10 −8 second, before it makes a “downward” transition and emits a photon. Assuming that the electron in hydrogen is in the state n = 2, how many revolutions about the nucleus are made before the electron “jumps” to the ground state? Solution The time period ‘T’ of electron in nth orbit is given by 1 n 3h 3 T= = f 4π 2 me 4 k 2

(2)3 (6.626 × 10−34 ) 4π 2 (9.109 × 10− 31 )(1.602 × 10−19 ) 4 (8.988 × 109 ) 2 T = 1.216 × 10 −15 s Number of revolutions completed by electron is given by Time in excited state 10 −8 = = 8.224 × 10 6 Revolutions Time period 1.216 × 10 −15 Example 4-7 Show that the energy of electron in nth for hydrogen atom is 13.6 given by E n = − 2 eV. n Solution The energy of electron in nth orbit is given by 2π 2 me 4 k 2 in joules En = − n2h2 T=

CHAPTER 04 ATOMIC STRUCTURE 99 2 3 2 2π me k En = − in eV n2h2 2π 2 (9.109 × 10 −31 )(1.602 × 10 −19 ) 3 (8.988 × 10 9 ) 2 En = − n 2 (6.626 × 10 −34 ) 2 13.6 En = − 2 in eV n Example 4-8 Calculate the binding energy of the hydrogen atom in the first excited state. Solution The energy of electron in nth orbit is given by 13.6 En = − 2 in eV n For first excited state we substitute n = 2, therefore 13.6 E2 = − 2 = −3.4 eV ( 2) Hence the binding energy of the hydrogen atom in the first excited state is 3.4 eV. Example 4-9 Calculate the energy of the electron for first two excited states of hydrogen atom. What will be the wavelength of emitted (or absorbed) radiation if an electron makes a transition between these two states? Solution (a) The energy of electron in nth orbit is given by 13.6 En = − 2 in eV n The energy of the first two excited states can be calculated by substituting n = 2 and n = 3 respectively. Hence 13.6 n = 2 , E2 = − 2 = −3.4 eV (2) 13.6 n = 3 , E3 = − 2 = −1.51 eV (3)

CHAPTER 04

ATOMIC STRUCTURE hc (b)Now E3 − E2 = hν =

100

λ

hc (6.626 × 10 −34 )(2.998 × 10 8 ) λ= = E3 − E 2 {−1.51 − (−3.4)}(1.602 × 10 −19 ) = 6.561 × 10 −7 m or 6561 Å

Example 4-10 Calculate the first, second and third excitation energies for hydrogen atom. Solution The energy of electron in nth orbit is given by 13.6 En = − 2 in eV n Now n = 1, E1 = −(13.6) /(1) 2 = −13.6 eV n = 2, E 2 = −(13.6) /(2) 2 = −3.4 eV n = 3, E3 = −(13.6) /(3) 2 = −1.51 eV n = 4,

E 4 = −(13.6) /(4) 2 = −0.85 eV

Let ε1, ε2 and ε3 be the first, second and third excitation energies of the hydrogen atom, then

ε1= E − E = −3.4 − (−13.6) = 10.2 eV ε2= E − E = −1.51 − (−13.6) = 12.09 eV ε3= E − E = −0.85 − (−13.6) = 12.75 eV 2

1

3

1

4

1

Example 4-11 Calculate the value of Rydberg constant for hydrogen atom. Solution The Rydberg constant is defined by 2π 2 k 2 me 4 R= ch3 2π 2 (8.988 ×109 ) 2 (9.109 × 10−31 )(1.602 × 10 −19 ) 4 R= (2.998 × 108 )(6.626 × 10−34 )3 R = 1.097 × 107 m −1

CHAPTER 04 ATOMIC STRUCTURE 101 Example 4-12 The wavelength of the first line of Lyman series of hydrogen atom is 1216 Ǻ. Calculate the wavelength of third line. B.U. B.Sc. 1990A Solution The wavelength of different members of the Lyman series of hydrogen atom is given by 1 1 1 = R  2 − 2  , n = 2,3,4,……. λn n  1 For the first member n = 2 , therefore 1 1  3R 1 = R 2 − 2  = 1216 2  4 1 1 -1 4 Ǻ R= == (3)1216) 912 For the third member n = 4 and formula for Lyman series becomes 1 1  15R 15 1 5 1 = R 2 − 2  = = × = λ3 4  16 16 912 4864 1 4864 λ3 = = 972.8 Ǻ 5 Example 4-13 The wavelength of the first line of Lyman series of hydrogen atom is 1216 Ǻ. Calculate the wavelength of 5th line. B.U. B.Sc. 1992A Solution The wavelength of different members of the Lyman series of hydrogen atom is given by 1 1 1 = R  2 − 2  , n = 2,3,4,……. λn n  1 For the first member n = 2 , therefore 4 1 -1 1 1  3R 1 or R = = Ǻ = R 2 − 2  = (3)1216) 912 1216 2  4 1

CHAPTER 04 ATOMIC STRUCTURE 102 For the fifth member n = 6 and formula for Lyman series becomes 1 1  35R 35 1 35 1 = R 2 − 2  = = × = λ3 36 912 32832 6  36 1 32832 λ3 = = 938 Ǻ 35 Example 4-14 The wavelength of the first member of Balmer series of the hydrogen atom is 6563 Å. Calculate the wavelength of the first member of Lyman series in the same spectrum. B.U. B.Sc. 2009A Solution The formula for Balmer series of hydrogen atom is 1 1 1 = R 2 − 2  , n = 3,4,5, . . . λ n  2 For first member of this series we substitute n = 3 1  1 1  5R = R 2 − 2  = λ1  2 3  36 36 λ1 = (1) 5R The Lyman series is given by 1 1 1 = R 2 − 2  , n = 2,3,4, . . . . λ n  1 For first member of this series we substitute n = 2 1 1  3R 1 = R 2 − 2  = λ1′ 2  4 1 4 λ1′ = (2) 3R Divide Eq(2) by Eq(1) λ1′  4  5R  5 =    = λ1  3R  36  27

CHAPTER 04 ATOMIC STRUCTURE 103 5λ1 5(6563) λ1′ = = = 1215.37 27 27 Example 4-15 Calculate the longest and shortest wavelength i.e. series limit of the Lyman series. Solution The Lyman series is given by 1 1 1 = R 2 − 2  , n = 2, 3, 4 . . . λ n  1 For longest wavelength, substitute n = 2 1 1  3R 1 = R 2 − 2  = λ 2  4 1

λ=

4 3R

4 = 121.5 ×10 −9 m = 1215 Å 7 3(1.097 ×10 ) For series limit, substitute n = ∞  1 1 1  1 = R  2 − 2  = R 1 − =R 2  λ n  1  (∞)  1 1 λ= = = 912 × 10 −10 m = 912 Ǻ 7 R 1.097 × 10 Example 4-16 Calculate the longest wavelength in Balmer series. Also find the value of the shortest wavelength. B.U. B.Sc. 1988S Solution The formula for Balmer series of hydrogen atom is 1 1 1 = R 2 − 2  , n = 3,4,5,…. λ n  2 Substitute n = 3 1 1  5R 1 = R 2 − 2  = λ 3  36 2

λ=

CHAPTER 04 ATOMIC STRUCTURE 104 36 36 λ= = = 6.563 × 10 − 7 m = 6563 Ǻ 7 5 R 5(1.097 × 10 ) Substitute n = ∞ in above formula 1 1  R 1 = R 2 − 2  = λ ∞  4 2 4 4 λ= = = 3.646 × 10 −7 m = 3646 Ǻ 7 R 1.097 × 10 The desired longest and shortest wavelengths are 6563 Ǻ and 3636 Ǻ respectively. Example 4-17 Calculate the longest and shortest wavelength i.e. series limit of the Paschen series. . B.U. B.Sc. 1991A Solution The formula for Paschen series of hydrogen atom is 1 1 1 = R 2 − 2  , n =4,5,6, . . . λ n  3 Substitute n = 4 1 1  7R 1 = R 2 − 2  = λ 4  144 3

144 144 = 7 R 7(1.097 × 107 ) λ = 1.875 ×10 −6 m = 1875 nm = 18750 Ǻ Substitute n = ∞ in above formula 1 1  R 1 = R 2 − 2  = λ ∞  9 3 9 9 λ= = R 1.097 × 107 λ = 8.204 ×10−7 m = 820.4nm = 8204 Ǻ The desired longest and shortest wavelengths are 1875 nm and 820.4 nm respectively.

λ=

CHAPTER 04 ATOMIC STRUCTURE 105 Example 4-18 Determine the largest wavelength when excited electron make transition to n = 4 level. K.U. B.Sc. 2001 Solution The wavelength of the emitted photon will; be the longest when electron jumps from n = 5 to n = 4. Hence 1 1  9R 1 = R 2 − 2  = λ  4 5  400

400 400 = 9 R 9(1.097 × 107 ) λ = 4.051 × 10−6 m = 4051 nm Note that this is the longest wavelength in Brackett series of Hydrogen atom. Example 4-19 Compute the first three wavelengths for the Paschen series of hydrogen. In what region of the spectrum do the lines of Paschen series lie? Solution The formula for Paschen series of hydrogen atom is 1 1 1 = R 2 − 2  , n =4,5,6, . . . λ n  3 1 1  7R 1 = R 2 − 2  = Substitute n = 4 , λ1 4  144 3 144 144 λ1 = = 7 R 7(1.097 ×107 ) λ1 = 1.875 × 10 −6 m = 1875 nm 1 1  16 R 1 Substitute n = 5 , = R 2 − 2  = λ2 5  225 3 225 225 = λ2 = 16 R 16(1.097 × 10 7 )

λ=

λ2 = 1.282 × 10 −6 m = 1282 nm

CHAPTER 04

ATOMIC STRUCTURE 106 1 1 R 1 Substitute n = 6 , = R 2 − 2  = λ3 6  12 3 12 12 λ3 = = = 1.094 × 10 −6 m = 1094 nm R 1.097 × 10 7 The wavelengths of desired lines are 1875 nm, 1282 nm and 1094 nm. These lines lie in the infrared region of the spectrum. Example 4-20 Find the wavelength of the spectral line corresponding to the transition in hydrogen atom from n = 6 to n = 3 state where R = 1.097 × 10 7 m -1 . B.U. B.Sc. 1991S Solution The desired line is the third line in Paschen series. The Paschen series is given by 1 1 1 = R 2 − 2  , n =4,5,6, . . . λ n  3 Substitute n = 6 1 1 R 1 = R 2 − 2  = λ 6  12 3 12 12 = R 1.097 × 107 λ = 1.0939 ×10 −6 m = 1093.9nm = 10939 Ǻ

λ=

Example 4-21 In the hydrogen atom, an electron experiences a transition from a state whose binding energy is 0.54 eV to another state whose excitation energy is 12.2 eV. (a) What are the quantum numbers for these states? (b) Compute the wavelength of the emitted photon. (c) To what series does this line belong? K.U. B.Sc. 2008 Solution The binding energy of the electron is given by 13.6 E n = 2 in eV n

CHAPTER 04 ATOMIC STRUCTURE (a)The above equation can be rewritten as 13.6 n= En

107

13.6 = 5. 0.54 The binding energy 0.54 eV corresponds to the state n = 5. Note that the binding energy of the state n = 1 is 13.6 eV. As the excitation energy is given as 10.2 eV, therefore E n = 13.6 − 10.2 = 3.4 eV Now E n = 0.54 eV, therefore n =

13.6 =2 3.4 i.e. the given excitation energy corresponds to state n = 2. hc (b) E2 − E5 = 3.4 − 0.54 = 2.86 eV =

Hence

n=

λ

hc λ= 2.86 eV (6.626 × 10 −34 )(2.998 × 108 ) λ= = 4.336 × 10 −7 m −19 (2.86)(1.602 × 10 ) λ = 4336 Å the wavelength of the emitted photon. (c) As the wavelength of the emitted photon is in the visible region, therefore it belongs to Balmer series of the hydrogen atom.

Example 4-22 A collection of hydrogen atoms in the ground state is illuminated with ultraviolet light of wavelength 59.0 nm. Find the kinetic energy of the emitted electron. Solution The energy of the incident photon is given by h c (6.626 × 10−34 )(2.998 × 108 ) E= = = 3.367 × 10−18 J −9 λ 59.0 × 10

CHAPTER 04 ATOMIC STRUCTURE 108 −18 3.367 × 10 E= eV = 21.0 eV 1.602 × 10−19 The energy needed to separate the electron from hydrogen atom is given by E1 = 13.6 eV

The kinetic energy of the emitted electron will be K = E − E1 = 21.0 − 13.6 = 7.4 eV

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ATOMIC STRUCTURE

109

4-2 HYDROGEN-LIKE ATOMS Example 4-23 How much energy is required to remove the electron from a singly ionized helium atom (He+1) in its ground state? Solution The energy of electron in nth orbit for hydrogen-like atom is given by

2π 2 me 4 k 2 Z 2 (13.6) Z 2 = − in eV n2h2 n2 Now n = 1 and Z = 2, therefore the desired energy is (13.6)(2) 2 E1 = − = −54.4 eV (1) 2 Example 4-24 Apply the Bohr’s theory to He+ and calculate for n = 1 (a) the radius, (b) the frequency of revolution, (c) the linear speed of the electron, (d) the total energy of the electron, (d) the angular momentum and (e) the ratio v/c, and (f) decide whether the classical treatment can be used or not. Solution (a) The radius of nth orbit of electron in hydrogen-like atom is given by n2h2 (5.287 × 10 −11 )n 2 rn = = Z 4π 2 k m Z e 2 Substitute n = 1, Z = 2 in above equation (5.287 × 10 −11 )(1) 2 r1 = = 2.644 × 10 −11 m 2 (b) The orbital frequency of the electron in hydrogen-like atom is given by 4π 2 m k 2 Z 2 e 4 fn = n3h3 4π 2 (9.109 × 10−31 )(8.988 × 109 )(2) 2 (1.602 × 10 −19 ) 4 f1 = (1)3 (6.626 × 10 −34 ) En = −

CHAPTER 04

ATOMIC STRUCTURE

110

16

= 2.631 × 10 Hz the desired frequency. (c) The velocity of electron in nth orbit of hydrogen-like atom is given by 2π k Z e 2 vn = nh 2π (8.988 × 109 )(2)(1.602 × 10−19 ) 2 v1 = (1)(6.626 × 10− 34 ) ν 1 = 4.375 ×106 m / s (d) The energy of electron in nth orbit for hydrogen-like atom is given by

2π 2 me 4 k 2 Z 2 (13.6) Z 2 = − in eV n2h2 n2 (13.6)(2) 2 E1 = − = −54.4 eV (1) 2 (e) The angular momentum of electron in first orbit is given by L1 = m v1 r1 = (9.109 × 10 −31 )(4.375 × 10 6 )(2.644 × 10 −11 ) = 1.054 × 10−34 kg m 2 s −1 En = −

v 4.375 × 10 6 = = 0.015 As the ratio (v/c) is small i.e. v c 2.998 × 10 8 is very small as compared with c, therefore classical treatment is applicable. Example 4-25 Determine the radius of the second Bohr orbit for doubly ionized Lithium. Solution The radius of nth orbit for Hydrogen-like atom is given by n 2r rn = 1 Z where r1 = 0.53 Å is the radius of the first Bohr orbit of Hydrogen atom. Hence (2) 2 (0.53) r2 = = 0.706 Å 3 (f)

CHAPTER 04 ATOMIC STRUCTURE 111 Example 4-26 How much energy is required to remove the electron from a doubly ionized lithium atom (Li+2) in its ground state? Solution The energy of electron in nth orbit for hydrogen-like atom is given by

2π 2 me 4 k 2 Z 2 (13.6) Z 2 En = − =− in eV n2h2 n2 Now n = 1 and Z = 3, therefore the desired energy is (13.6)(3) 2 E1 = − = −122.4 eV (1) 2 Example 4-27 (a) Compute the first and second excitation potentials for singly ionized helium atom. (b) What wavelengths are emitted when the He+ returns to the ground state from these excited states? Solution The energy of electron in nth orbit for hydrogen-like atom is given by 2π 2 me 4 k 2 Z 2 (13.6) Z 2 = − in eV n2h2 n2 (13.6)(2) 2 54.4 =− = − 2 eV for He+ 2 n n 54.4 Now n = 1 , E1 = − 2 = −54.4 eV (1) 54.4 n = 2 , E 2 = − 2 = −13.6 eV ( 2) 54.4 n = 3 , E3 = − 2 = −6.04 eV (3) En = −

Let ε1 and ε2 be the first and second excitation energies of the He+ atom, then

CHAPTER 04

ε1= E ε2= E

2

ATOMIC STRUCTURE

112

− E1 = −13.6 − (−54.4) = 40.8 eV

− E1 = −6.04 − (−54.4) = 48.46 eV The desired excitation potentials are 40.8 V and 48.46 V. hc (6.626 × 10 −34 )(2.998 × 10 8 ) (b) Now ε1 = hc or λ1 = = ε1 (40.8)(1.602 × 10 −19 ) λ1 3

λ1 = 3.039 ×10 −8 m ≅ 304 Å

ε2 = hc λ2

or λ2 =

hc

ε 21

=

(6.626 × 10 −34 )(2.998 × 10 8 ) (48.46)(1.602 × 10 −19 )

λ1 = 2.559 × 10 −8 m ≅ 256 Å Example 4-28 Determine the wavelength of the first two lines of singly ionized helium that corresponds to the first two lines of the Balmer series. Solution The formula for Balmer series of hydrogen-like atom is 1 1 1 = R Z2 2 − 2  , n = 3,4,5,….. λ n  2 For singly ionized helium Z = 2 and above equation becomes 1 1 1 = 4 R  2 − 2  , n = 3,4,5,….. λ n  2 The desired wavelengths are calculated by substituting n = 3 and n = 4 respectively in above relation. Now 1  1 1  5R = 4R  2 − 2  = λ1 2 3  9 9 9 λ1 = = = 1.641× 10 −7 m = 1641 Å 7 5 R 5(1.097 × 10 ) 1 1  3R 1 = 4R  2 − 2  = λ2 4  4 2 4 4 λ2 = = = 1.215 × 10 −7 m = 1215 Å 3R 3(1.097 × 10 7 )

CHAPTER 04 ATOMIC STRUCTURE 113 Example 4-29 Calculate the wavelengths of first two lines of the Balmer series of triply ionized Beryllium (Z = 4). Solution The formula for Balmer series of Hydrogen-like atom is 1 1 1 = R Z2 2 − 2  , n = 3,4,5,….. λ n  2 For present case Z = 4 , therefore 1 1 1 = 16 Z 2  2 − 2  , n = 3,4,5,….. λ n  2

 1 1  20 R = 16 R  2 − 2  = λ1 9 2 3  9 9 λ1 = = = 4.10 × 10−8 m = 41.0 nm 7 20 R 20(1.097 × 10 ) 1 1 1 n = 4, = 16 R  2 − 2  = 3 R λ2 4  2 1 1 λ2 = = = 3.04 × 10−8 m = 30.4 nm 7 3 R 3(1.097 × 10 ) Hence the wavelengths of the first two lines of the Balmer series of triply ionized Beryllium are 41.0 nm and 30.4 nm respectively. Example 4-30 An atom of tungsten has all of its electrons removed except one. (a) Calculate the ground-state energy for this one remaining electron. (b) Calculate the energy and wavelength of the radiation emitted when this electron makes a downward transition from n = 2 to n = 1. (c) In what portion of the electromagnetic spectrum is this photon? Solution (a)The energy of electron in nth orbit for hydrogen-like atom is given by Now n = 3,

1

CHAPTER 04 2

ATOMIC STRUCTURE 4

2

2

2π me k Z (13.6) Z =− 2 2 n h n2 For tungsten Z = 74, therefore En = −

114

2

in eV

(13.6)(74) 2 7.44736 × 10 4 74.5 = − eV ≅ − 2 keV 2 2 n n n E1 = −74.5 keV the desired ground state energy. (b) When electron makes a downward transition from n = 2 to n = 1, the energy ‘E’ of the emitted photon is given by En = −

1 1 E = E 2 − E1 = −74.5 2 − 2  keV = 55.875 keV ≅ 55.9 keV 1  2 The wavelength of the emitted photon is calculated from the relation hc (6.626 × 10−34 )(2.998 × 108 ) λ= = E (55.875 × 103 )(1.602 × 10−19 ) λ = 2.219 × 10 −11 m ≅ 0.222 Å (c) The emitted photon belongs to X-rays.

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ATOMIC STRUCTURE

115

4-3 THE FRANK-HERTZ EXPERIMENT Example 4-31 In Franck-Hertz experiment the ionization potential of sodium is 5.13 V. Calculate the velocity of the electron. Solution The kinetic energy of the electron is given by 1 m v 2 = Ve 2

v=

2Ve = m

2(5.13)(1.602 × 10 −19 ) = 1.343 × 10 6 m s −1 − 31 (9.109 × 10 )

Example 4-32 What will be the minimum energy of electron to excite 2536 Å spectral line of mercury in a Franck-Hertz experiment? Solution The minimum kinetic energy of electron to excite a spectral line of wavelength λ is given by hc (6.626 × 10 −34 )(2.998 × 10 9 ) ( K .E.) min = = λ 2536 × 10 −10 = 7.833 × 10 −19 J or 4.890 eV

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ATOMIC STRUCTURE

116

4-4 ANGULAR MOMENTUM Example 4-33 Calculate the value of Bohr magneton. Solution The Bohr magneton is defined as eh eh µB = or 4π me 2me

µB =

(1.602 × 10 −19 )(6.626 × 10 −34 ) = 9.273 × 10 −24 J T −1 −31 4π (9.109 × 10 )

The accepted value of Bohr magneton is 9.274 × 10 −24 J T −1 or 5.778 × 10 −5 eV T −1 . Example 4-34 Calculate the magnitude of the orbital angular momentum of an electron in a state with l = 4 . Solution The magnitude of orbital angular momentum is given by

L = l(l + 1)h = 4(4 + 1) (1.055 × 10 −34 ) = 4.718 × 10 −34 J s Example 4-35 Write down the quantum numbers for all the hydrogen atom states belonging to subshells for which n = 4 and l = 3 . Solution Now n = 4 and l = 3 . As ml = 0, ±1 ,±2,........... ± l therefore the desired magnetic quantum numbers are ml = −3, −2, −1, 0, 1, 2, 3 1 The spin quantum number is given by m S = ± . 2

CHAPTER 04 ATOMIC STRUCTURE 117 Example 4-36 A hydrogen state is known to have the quantum number l = 3 . What are possible n, ml and m S . Solution The principal quantum number ‘n’ must be greater than ‘ l ’ which is 3 for present case. Hence n = 4 and the magnetic quantum number ‘ ml ’ value are as under ml = −3, −2, −1, 0, 1, 2, 3

1 The spin quantum number is given by m S = ± . 2 Example 4-37 If an electron in a hydrogen atom is in a state with l = 5 , what is the smallest possible angle between L and LZ . Solution The minimum angle occurs when ml = l and is given by cos[θ min ] =

cos[θ min ] =

LZ = L

5 5(5 + 1)

l h l(l + 1) h

=

l l(l + 1)

= 0.91287

θ min = 24.10 or 2405′42′′ Example 4-38 Calculate and tabulate for a hydrogen atom in a state l = 3 , the allowed values of LZ , µ Z and θ . Find the magnitude of L and µ . Solution The corresponding values of magnetic quantum number ml for l = 3 are ml = −3, −2, −1, 0, 1, 2, 3 .The desired values are calculated and tabulated below. Here we have employed the following values

CHAPTER 04 ATOMIC STRUCTURE h = 1.055 × 10 −34 J s and µ B = 9.274 × 10 −24 J / T

ml

L Z = ml h

µ Z = − ml µ B

kg m 2 / s

J /T

118

 ml  θ = cos −1    l(l + 1) 

−34

− 3.165 × 10 2.782 × 10 −23 150.0 0 -3 -2 − 2.110 × 10 −34 1.855 × 10 −23 125.3 0 -1 − 1.055 × 10 −34 9.274 × 10 −24 106.8 0 0 0 0 90.0 0 1 1.055 × 10 −34 − 9.274 × 10 −24 73.2 0 2 2.110 × 10 −34 − 1.855 × 10 −23 54.7 0 3 3.165 × 10 −34 − 2.782 × 10 −23 30.0 0 The magnitude of orbital angular momentum is given by

L = l(l + 1)h = 3(3 + 1) (1.055 × 10 −34 ) = 3.655 × 10 −34 J s The magnitude of dipole moment µ is given by

µ = µ B l(l + 1) = (9.274 × 10 −24 )( 3(3 + 1) ) = 3.213 × 10 23 J / T

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4-5 THE ZEEMAN EFFECT Example 4-39 A hydrogen atom gas is placed in intense magnetic field of 0.8 tesla. Compute the separation between two consecutive states (Normal Zeeman Effect) when λ = 5000 Å. Solution The wavelength separation is given by µ Bλ2 eBλ2 eh eh ∆λ = = B where µ B = or hc 4π me 2me 4π me c

∆λ =

(9.274 × 10 −24 )(0.8)(5000 × 10 −10 ) 2 (6.626 × 10 −34 )(2.998 × 10 8 )

∆λ = 9.337 × 10 −12 m or 0.0937 Å

Example 4-40 What will be the separation, in Å, due to normal Zeeman splitting of 4916 Å line in the mercury spectrum in a magnetic field of 0.5 tesla? Solution The wavelength separation is given by µ B Bλ2 (9.274 × 10 −24 )(0.5)(4916 × 10 −10 ) ∆λ = = hc (6.626 × 10 −34 )(2.998 × 10 8 ) ∆λ = 5.64 × 10 −12 m = 5.64 × 10 −2 Å

Example 4-41 Calculate the wavelengths of the normal Zeeman Effect triplet of a spectral line 5500 Å placed in a magnetic field of 10 tesla. Solution The wavelength separation is given by µ B Bλ2 (9.274 × 10 −24 )(10)(5500 × 10 −10 ) 2 ∆λ = = hc (6.626 × 10 −34 )(2.998 × 10 8 )

CHAPTER 04 ATOMIC STRUCTURE 120 −10 ∆λ = 1.412 × 10 m = 1.412 Å The wavelengths of Zeeman triplet are λ1 = λ − ∆λ = 5500 − 1.412 = 5498.588 Å λ2 = λ = 5500 Å λ3 = λ + ∆λ = 5500 + 1.412 = 5501.412 Å Example 4-42 The Zeeman components of a 500 nm spectral line are 0.0116 nm apart when the magnetic field is 1.00 tesla. Find the ratio of e/m of the electron from this data. Solution The wavelength separation is given by eBλ2 ∆λ = 4π me c

or

e 4π c(∆λ ) = me Bλ2 e 4π (2.998 × 10 8 )(0.0116 × 10 −9 ) = me (1)(500 × 10 −9 ) 2 e = 1.733 × 1011 C / kg me

Example 4-43 A spectral line having a wavelength of 5500 Å shows a normal Zeeman splitting of 1.1 × 10 −2 Å. What is the magnitude of the magnetic field causing the splitting? Solution The wavelength separation is given by µ B Bλ2 ∆λ = hc hc(∆λ ) (6.626 ×10 −34 )(2.998 ×108 ){(1.1×10−2 )(10 −10 )} or B = = µ B λ2 (9.274 ×10 −24 )(5500 ×10 −10 ) 2 B = 0.078 tesla

CHAPTER 04 ATOMIC STRUCTURE 121 Example 4-44 Find the minimum magnetic field needed for the Zeeman Effect to be observed in a spectral line of 400 nm wavelength when a spectrometer whose resolution is 0.010 nm is used. Solution The wavelength separation is given by µ Bλ2 ∆λ = B hc hc(∆λ ) or B= 2

µBλ

(6.626 ×10 −34 )(2.998 ×108 ){(0.010 ×10 −9 )} (9.274 ×10 − 24 )(400 ×10 −9 ) 2 B = 1.339 tesla B=

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CONCEPTUAL QUESTIONS (1) What is meant by a stationary orbit in Bohr’s theory of Hydrogen atom? Answer: - A stationary or allowed orbit is one in which electron does not radiate any energy i.e. the total energy of the electron remains constant. (2) What should be the angular momentum of the electron moving in an allowed orbit? Answer: - The angular momentum of an electron should be an integral multiple of h = (h / 2 π ) while moving in an allowed orbit i.e. L = n h . (3) How many electronic orbits are there in a Hydrogen atom? Answer: - Theoretically there are an infinite number of orbits in a Hydrogen atom. As there is only one electron in the Hydrogen atom, therefore one of these orbits will be occupied at one time. The remaining orbits will be vacant. (4) How is the spacing of adjacent electronic energy levels of an atom affected by the value of n? Answer: - The spacing of adjacent electronic energy levels of an atom decreases as n increases. (5) What is indicated by the existence of sharp lines in the spectrum of Hydrogen atom? Answer: - It indicates that the electron is moving in a given orbit with certain frequency only. OR It indicates the presence of discrete energy levels in the Hydrogen atom. (6) Which transition of the electron in Hydrogen atom will emit highest energy photon? Answer: - The transition of the electron from n = ∞ to n = 1 will emit maximum energy photon. The energy of this photon is E∞ − E1 = (0) − (−13.6 eV ) = 13.6 eV .

CHAPTER 04 ATOMIC STRUCTURE 123 (7) The total energy of the electron in Hydrogen atom is negative. What is the physical significance of negative sign? Answer: - It reflects the fact that the electron is bound to the nucleus. An energy of 13.6 eV is needed to separate the electron from the nucleus of hydrogen atom. (8) Can a Hydrogen atom absorb a photon whose energy exceeds its biding energy? Answer: - Yes. The excess energy will appear as kinetic energy of the detached electron of Hydrogen atom. (9) Which series in the emission spectrum of Hydrogen atom has the highest frequencies? In what part of the spectrum are these lines? Answer: - Lyman series. The spectral lines of Lyman series are in the ultraviolet region. (10) Any series of atomic Hydrogen yet to be discovered will probably found to be in what region of the spectrum? Answer: - In infrared region of the spectrum. (11) If an electron moves to a larger orbit, does its total energy increase or decrease? Does its kinetic energy increase or decrease? Answer: - The total energy of an orbiting electron is given by k Z e2 E=− 2 r As r increases E becomes less negative and therefore increases. The orbital kinetic energy of the electron is given by k Z e2 K= 2r It is clear that K decreases with increase in r. (12) Under what conditions an electron moving in the orbit will have de Broglie wavelength? Answer: -The electron orbit (i.e. the circumference of circular orbit) must contain an integral number of de Broglie wavelengths.

CHAPTER 04 ATOMIC STRUCTURE 124 (13) What are two essential features of the vector atom model? Answer: - (a) Space quantization and (b) Hypothesis of spinning electron. (14) What is minimum energy needed to create an Electron positron pair? Answer: - The minimum energy needed to create an electronpositron pair is given by 2 m0c 2 = 1.022 MeV . (15) What is the wavelength of the photon corresponding to the threshold energy needed to produce electron-positron pair i.e. 1.022 MeV? Answer: - 0.012 Å

CHAPTER 04

(1) (2) (3)

(4) (5)

(6)

(7)

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ADDITIONAL PROBLEMS Compute the first three wavelengths of the Bracket series of hydrogen. The binding energy of electron in a certain state of hydrogen atom is 0.85 eV. What is the corresponding quantum number? A photon of energy 12.1 eV absorbed by a hydrogen atom, originally in the ground state, raises the atom to an excited state. What is the quantum number of this state? Determine the speed of electron in the n = 8 orbit of hydrogen atom. Use this value to find the corresponding de Broglie wavelength. The lifetime of an excited state is about 10 −8 second. Compute how many revolutions an electron of hydrogen atom in the excited state n = 4 will make before jumping in the ground state? Which of the spectral lines of the Bracket series is closest in wavelength to the first spectral line of the Paschen series of hydrogen atom? What is difference in their wavelengths? Find the radius of the smallest Bohr orbit in doubly ionized lithium ( Li + + ). What will be the energy of electron in this orbit? Answers −6

(1) 4.051 × 10 m, 2.625 × 10 −6 m and 2.166 × 10 −6 m (2) 4 (3) 3 (4) 2.734 × 10 5 m s −1 , 2661 Å (5) 1.04 × 10 7 Revolutions (6) Fifth line (n = 9) (7) 0.18 Å , -122 eV