(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2,
Question:

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the = 1, 2, and 3 levels.

(b) Calculate the orbital period in each of these levels.

Solution:

(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, νis given by the relation,

$v_{1}=\frac{e^{2}}{n_{1} 4 \pi \in_{0}(h / 2 \pi)}=\frac{e^{2}}{2 \in_{0} h}$

Where,

= 1.6 × 10−19 C

0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2

h = Planck’s constant = 6.62 × 10−34 Js

$\therefore v_{1}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}$

$=0.0218 \times 10^{8}=2.18 \times 10^{6} \mathrm{~m} / \mathrm{s}$

For level n2 = 2, we can write the relation for the corresponding orbital speed as:

$v_{2}=\frac{e^{2}}{n_{2} 2 \epsilon_{0} h}$

$=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}$

$=1.09 \times 10^{6} \mathrm{~m} / \mathrm{s}$

And, for n3 = 3, we can write the relation for the corresponding orbital speed as:

$v_{3}=\frac{e^{2}}{n_{3} 2 \epsilon_{0} h}$

$=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{3 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}$

$=7.27 \times 10^{5} \mathrm{~m} / \mathrm{s}$

Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 106 m/s, 1.09 × 106 m/s, 7.27 × 105 m/s respectively.

(b) Let T1 be the orbital period of the electron when it is in level n1 = 1.

Orbital period is related to orbital speed as:

$T_{1}=\frac{2 \pi r_{1}}{v_{1}}$

Where,

r1 = Radius of the orbit

$=\frac{n_{1}{ }^{2} h^{2} \in_{0}}{\pi m e^{2}}$

h = Planck’s constant = 6.62 × 10−34 Js

e = Charge on an electron = 1.6 × 10−19 C

= Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2

m = Mass of an electron = 9.1 × 10−31 kg

$\therefore T_{1}=\frac{2 \pi r_{1}}{v_{1}}$

$=\frac{2 \pi \times(1)^{2} \times\left(6.62 \times 10^{-34}\right)^{2} \times 8.85 \times 10^{-12}}{2.18 \times 10^{6} \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^{2}}$

$=15.27 \times 10^{-17}=1.527 \times 10^{-16} \mathrm{~s}$

For level $n_{2}=2$, we can write the period as:

$T_{2}=\frac{2 \pi r_{2}}{v_{2}}$

Where,

r2 = Radius of the electron in n2 = 2

$=\frac{\left(n_{2}\right)^{2} h^{2} \in_{0}}{\pi m e^{2}}$

$\therefore T_{2}=\frac{2 \pi r_{2}}{v_{2}}$

$=\frac{2 \pi \times(2)^{2} \times\left(6.62 \times 10^{-34}\right)^{2} \times 8.85 \times 10^{-12}}{1.09 \times 10^{6} \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^{2}}$

$=1.22 \times 10^{-15} \mathrm{~s}$

And, for level n3 = 3, we can write the period as:

$T_{3}=\frac{2 \pi r_{3}}{v_{3}}$

Where,

r3 = Radius of the electron in n3 = 3

$=\frac{\left(n_{3}\right)^{2} h^{2} \in_{0}}{\pi m e^{2}}$

$\therefore T_{3}=\frac{2 \pi r_{3}}{v_{3}}$

$=\frac{2 \pi \times(3)^{2} \times\left(6.62 \times 10^{-34}\right)^{2} \times 8.85 \times 10^{-12}}{7.27 \times 10^{5} \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^{2}}$

$=4.12 \times 10^{-15} \mathrm{~s}$

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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