Question:

An electron in a hydrogen atom excites from $ n = 2 $ to $ n = 4 $. What is the change in angular momentum? 
(Planck's constant $ h = 6.64 \times 10^{-34} \, \text{J s} $)

Show Hint

In Bohr's model, angular momentum \( L = n\hbar \), where \( \hbar = \frac{h}{2\pi} \).
Updated On: May 20, 2025
  • \( 2.11 \times 10^{-34} \, \text{Js} \)
     

  • \( 1.05 \times 10^{-34} \, \text{Js} \)
     

  • \( 0.57 \times 10^{-34} \, \text{Js} \)
     

  • \( 4.22 \times 10^{-34} \, \text{Js} \)
     

Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Angular momentum of electron in \( n^{\text{th}} \) orbit: \[ L_n = n \cdot \frac{h}{2\pi} \] Change in angular momentum: \[ \Delta L = L_4 - L_2 = \left(4 - 2\right) \cdot \frac{h}{2\pi} = 2 \cdot \frac{6.64 \times 10^{-34}}{2\pi} = \frac{6.64 \times 10^{-34}}{\pi} \approx 2.11 \times 10^{-34}\, \text{Js} \]

Was this answer helpful?
0
0