Class 12 Physics - Chapter Atoms NCERT Solutions | (a) Using the Bohr’s model calcula
(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.
(b) Calculate the orbital period in each of these levels.
(a) Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation,
ν 1 = e2/n14πϵ0(h/2π) = e2/2ϵ0h
Where, e = 1.6 × 10−19 C
ϵ0 = Permittivity of free space = 8.85 × 10-12 N−1 C2 m−2
h = Planck’s constant = 6.62 × 10−34 Js
∴ ν1 = (1.6x10-19)2/2x8.85x10-12x6.62x10-34 = 0.0218 x 108 = 2.18 x 106 m/s
For level n2 = 2, we can write the relation for the corresponding orbital speed as:
ν2 = e2/n22ϵ0h = (1.6x10-19)2/2x2x8.85x10-12x6.62x10-34 = 1.09 x 106 m/s
And, for n3 = 3, we can write the relation for the corresponding orbital speed as:
ν3 = e2/n32ϵ0h = (1.6x10-19)2/3x2x8.85x10-12x6.62x10-34 = 7.27 x 105 m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 10 6 m/s, 1.09 × 10 6 m/s, 7.27 × 10 5 m/s respectively.
(b) Let T 1 be the orbital period of the electron when it is in level n1 = 1.
Orbital period is related to orbital speed as:
T1 = 2πr1/ν 1
Where, r1 = Radius of the orbit
= n12h2ϵ0/πme2
h = Planck’s constant = 6.62 × 10−34 Js
e = Charge on an electron = 1.6 × 10−19 C
ϵ0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
m = Mass of an electron = 9.1 × 10−31 kg
∴ T1 = 2πr1/ν 1
= (2πx(1)2x(6.62x10-34)2x8.85x10-12)/2.18x106xπx9.1x10-31x(1.6x10-19)2
= 15.27x10-17 = 1.527x10-16 s
For level n 2 = 2, we can write the period as:
T2 = 2πr2/ν 2
Where, r2 = Radius of the electron in n2 = 2
= (n2)2h2ϵ0/πme2
∴ T2 = 2πr2/ν2
= (2πx(2)2x(6.62x10-34)2x8.85x10-12)/1.09 x 106 x π x 9.1 x 10-31 x (1.6 x 10-19)2
= 1.22 x 10-15 s
And, for level n 3 = 3, we can write the period as:
T3 = 2πr3/ν 3
Where, r 3 = Radius of the electron in n 3 = 3
= (n3)2h2ϵ0/πme2
∴ T3 = 2πr3/ν 3
= (2πx(3)2x(6.62x10-34)2x8.85x10-12)/7.27 x 105 x π x 9.1 x 10-31 x (1.6 x 10-19)2
= 4.12 x 10-15 s
Hence, the orbital period in each of these levels is 1.52 × 10 −16 s, 1.22 × 10 −15 s, and 4.12 × 10 −15 s respectively.
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[Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]
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