Question

When an electron in a hydrogen atom is excited from $n=4$ to $n=5$, the change in angular momentum of the electron is,
$($plank's constant, $h=6.6\times {10}^{-34}\text{J s})$

• A. $4.16\times {10}^{-34}\text{J s}$
• B. $3.32\times {10}^{-34}\text{J s}$
• C. $1.05\times {10}^{-34}\text{J s}$
• D. $2.08\times {10}^{-34}\text{J s}$
  

Answer 1

The correct option is C $1.05\times {10}^{-34}\text{J s}$

Change in the angular momentum is given as,
$\Delta L=L_2-L_1=\frac{n_{2}h}{2\pi }-\frac{n_{1}h}{2\pi }$
$\Delta L=\frac{h}{2\pi }(n_2-n_1)$
$\Delta L=\frac{6.6\times {10}^{-34}}{2\times 3.14}(5-4)=1.05\times {10}^{-34}\text{J s}$

Hence, $\left(C\right)$ is the correct answer.
$$\begin{array}{c}\text{Why this question?}\\ \text{This question is based on angular momentum}\left(Boh{r}^{\prime }ssecondpostulates\right)\text{, the question will give a clear understanding on how to calculate difference in angular momentum.}\end{array}$$