Question

An electron in the hydrogen atom excites from \(2^{nd}\) orbit to \(4^{th}\) orbit, then the change in angular momentum of the electron is \((\text{Planck's constant }h=6.64\times10^{-34}\,\text{J-s})\)

• A. \(2.11\times10^{-34}\,\text{J-s}\)
• B. \(1.05\times10^{-34}\,\text{J-s}\)
• C. \(0.57\times10^{-34}\,\text{J-s}\)
• D. \(4.22\times10^{-34}\,\text{J-s}\)

Hint:

In Bohr's atomic model, angular momentum is quantized as \[ L_n=\frac{nh}{2\pi} \] So, change in angular momentum is \[ \Delta L=\frac{(n_f-n_i)h}{2\pi} \]

Answer 1

The Correct Option is A

Step 1: State Bohr's quantization condition.
According to Bohr's second postulate, the angular momentum of an electron in a stationary orbit is quantized: \[ L_n = \frac{nh}{2\pi} \] where $n$ is the principal quantum number and $h$ is Planck's constant.
Step 2: Write the angular momentum in the initial orbit.
The electron starts in the 2nd orbit ($n_i = 2$): \[ L_i = \frac{2h}{2\pi} = \frac{h}{\pi} \]
Step 3: Write the angular momentum in the final orbit.
The electron moves to the 4th orbit ($n_f = 4$): \[ L_f = \frac{4h}{2\pi} = \frac{2h}{\pi} \]
Step 4: Calculate the change in angular momentum.
\[ \Delta L = L_f - L_i = \frac{4h}{2\pi} - \frac{2h}{2\pi} = \frac{(4-2)h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \]
Step 5: Substitute the numerical value of $h$.
Given $h = 6.64 \times 10^{-34}\,\text{J s}$ and $\pi \approx 3.14$: \[ \Delta L = \frac{6.64 \times 10^{-34}}{3.14} \approx 2.11 \times 10^{-34}\,\text{J s} \]
Step 6: State the final answer.
The change in angular momentum of the electron as it moves from the 2nd to the 4th orbit is: \[ \boxed{\Delta L = 2.11 \times 10^{-34}\,\text{J s}} \]