Question

The velocity of an electron in the nth orbit of hydrogen atom bears the ratio 1:411 to the velocity of light. The number of coloured lines formed when electron jumps from (n+3) state is:
A. 4
B. 3
C. 5
D. 6

Answer 1

Hint: We need to understand Bohr’s atomic model to find the number of coloured lines when an electron jumps in a hydrogen atom. We are to discuss the hydrogen spectra given by Bohr’s atomic model. This theory of his is known as the planetary model where electrons move in fixed paths around the nucleus called orbits whose energy is fixed. This planetary model explains the emissions of a hydrogen atom.

Complete step by step answer:
The velocity of the electron in a hydrogen atom revolving around in the 1st orbit is calculated to be $$2.18\times {10}^{6}m/s$$. The velocity of the electron in the nth orbit is given by $v_n=2.18\times {10}^6\times {\displaystyle \frac{z}{n}}$,where $z$is the atomic number of hydrogen atom which is equal to 1.
Hence, $v_n=2.18\times {10}^6\times {\displaystyle \frac{1}{n}}$ ………. (1)
Given: the ratio velocity of an electron in the nth orbit to the velocity of light ($v$) is 1:411.
We know that the speed of light in a vacuum is $3\times {10}^{8}m/s$. Therefore, ${\displaystyle \frac{v_n}{v}}={\displaystyle \frac{1}{411}}$
Or, $v_n={\displaystyle \frac{1}{411}}\times 3.8\times {10}^8$
$=7.299\times {10}^{5}m/s$
Putting this value of $v_n$ in equation we have,
${\displaystyle \frac{7.299\times {10}^5}{2.18\times {10}^6}}={\displaystyle \frac{1}{n}}$
Or, $n={\displaystyle \frac{2.18\times {10}^6}{7.299\times {10}^5}}$
$=2.986$
$\approx 3$ (since n cannot be a decimal number).
So, it can be concluded that the electron jumps from the orbit n=3 to the orbit n+3=6 orbit which is from the 3rd orbit to the 6th orbit ($n_1=3,n_2=6$)
Therefore, the number of coloured lines formed when an electron jumps from $n_1$to $n_2$is,
${\displaystyle \frac{(n_2-n_1)(n_2-n_1+1)}{2}}={\displaystyle \frac{\left(6-3\right)\left(6-3+1\right)}{2}}={\displaystyle \frac{3\times 4}{2}}=6$

So, the correct answer is Option D.

Note: It must be noted to keep the value of n always as a positive integer. Further, the velocity of the electron in the first orbit of a hydrogen atom is calculated using the formula
$v={\displaystyle \frac{nh}{2mr\pi }}$
where n is the number of orbits, h is the Planck’s constant, m is the mass of an electron, r is the radius of the orbit. This formula is derived from the de Broglie Equation of angular momentum.