Question

When an electron in a hydrogen atom is excited from its $4^{th}$ to $5^{th}$ stationary orbit, the change in angular momentum of the electron is $(h=6.6\times {10}^{-34}\text{Js})$

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

Answer 1

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

According to Bohr's hypothesis, angular momentum of the stationary orbits is given by,

$L=\frac{nh}{2\pi }$

The change in angular momentum is given by,

$\Delta L=L_2-L_1=\frac{n_{2}h}{2\pi }-\frac{n_{1}h}{2\pi }=\frac{h}{2\pi }(n_2-n_1)$

$=\frac{6.6\times {10}^{-34}}{2\times 3.14}(5-4)=1.05\times {10}^{-34}\text{Js}$

Hence, option $\left(C\right)$ is correct.

Why this Question? The question gives a clear understanding of how to calculate the difference in angular momentum.