Question

A hydrogen atom in ground state absorbs $10.2\text{eV}$ of energy. The orbital angular momentum of the electron is increased by

• A. $3.16\times {10}^{-34}\text{Js}$
• B. $1.05\times {10}^{-34}\text{Js}$
• C. $2.11\times {10}^{-34}\text{Js}$
• D. $4.22\times {10}^{-34}\text{Js}$

Answer 1

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

The energy of an electron present in ground state of a hydrogen atom will be,

$E=-13.6\text{eV}$

After absorbing $10.2\text{eV}$, the energy of the electron becomes

$E=-13.6+10.2=-3.4\text{eV}$

$E=-3.4\text{eV}$ corresponds to the energy of the first excited state $(n=2)$

Thus, an electron jumps from ground state $(n_1=1)$ to first excited state$(n_2=2)$

According to Bohr's postulate, the angular momentum of the stationary states is given by,

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

$\therefore$ Increase in Angular momentum is given by

$\Delta L=\frac{n_{2}h}{2\pi }-\frac{n_{1}h}{2\pi }$

$\Delta L=\frac{2h}{2\pi }-\frac{h}{2\pi }$

$\Rightarrow \Delta L=\frac{h}{2\pi }$

$\Rightarrow \Delta L=\frac{6.6\times {10}^{-34}}{2\times 3.14}$

$\Rightarrow \Delta L=1.05\times {10}^{-34}\text{Js}$

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

Note: It's a very common question in both JEE & NEET.