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: Use Bohr's quantization condition.
According to Bohr's model, angular momentum of an electron in the \(n^{th}\) orbit is \[ L_n=\frac{nh}{2\pi} \] where, \[ n=\text{principal quantum number} \] and \[ h=\text{Planck's constant} \]

Step 2: Write angular momentum in the initial orbit.
The electron is initially in the \(2^{nd}\) orbit.
So, \[ n_i=2 \] Hence, \[ L_i=\frac{2h}{2\pi} \]

Step 3: Write angular momentum in the final orbit.
The electron goes to the \(4^{th}\) orbit.
So, \[ n_f=4 \] Hence, \[ L_f=\frac{4h}{2\pi} \]

Step 4: Calculate change in angular momentum.
\[ \Delta L=L_f-L_i \] \[ \Delta L=\frac{4h}{2\pi}-\frac{2h}{2\pi} \] \[ \Delta L=\frac{(4-2)h}{2\pi} \] \[ \Delta L=\frac{2h}{2\pi} \] \[ \Delta L=\frac{h}{\pi} \]

Step 5: Substitute the value of \(h\).
Given, \[ h=6.64\times10^{-34}\,\text{J-s} \] Therefore, \[ \Delta L=\frac{6.64\times10^{-34}}{\pi} \] Using \[ \pi \approx 3.14 \] \[ \Delta L=\frac{6.64\times10^{-34}}{3.14} \] \[ \Delta L=2.11\times10^{-34}\,\text{J-s} \]

Step 6: Final conclusion.
Hence, the change in angular momentum is \[ \boxed{2.11\times10^{-34}\,\text{J-s}} \]