Question

If the angular momenta of electrons in two orbits of hydrogen atom are \( \frac{h}{\pi} \) and \( \frac{1.5h}{\pi} \), then the ratio of velocities of electrons in these two orbits is
(h – Planck’s constant)

• A. \( 3 : 2 \)
• B. \( 3 : 4 \)
• C. \( 9 : 4 \)
• D. \( 1 : 3 \)

Hint:

In Bohr's model, angular momentum is quantized and related to orbit number \( n \). Use \( v \propto \frac{1}{n} \) and \( L \propto n \) to find ratios effectively.

Answer 1

The Correct Option is A

The angular momentum for hydrogen atom is given by: \[ L = mvr \] From Bohr's quantization condition: \[ L = n\hbar = \frac{nh}{2\pi} \] In this question, we are given: \[ L_1 = \frac{h}{\pi}, \quad L_2 = \frac{1.5h}{\pi} \] Since \( L = mvr \) and for hydrogen-like atoms \( r \propto n^2 \) and \( v \propto \frac{1}{n} \), we can also say: \[ v \propto \frac{L}{r} \propto \frac{L}{n^2} \] Let’s use the definition \( L = mvr \Rightarrow v = \frac{L}{mr} \), and from Bohr’s model \( r \propto n^2 \), and since \( L \propto n \), then: \[ v \propto \frac{n}{n^2} = \frac{1}{n} \] Hence, \[ v_1 : v_2 = \frac{L_1}{r_1} : \frac{L_2}{r_2} = \frac{L_1 / m}{n_1^2} : \frac{L_2 / m}{n_2^2} \] Now, since the angular momenta are in the ratio: \[ \frac{L_2}{L_1} = \frac{1.5h/\pi}{h/\pi} = \frac{3}{2} \Rightarrow n_2 : n_1 = \frac{3}{2} \Rightarrow v_1 : v_2 = \frac{1}{1} : \frac{1}{(3/2)} = 3 : 2 \]