Answer:
(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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