JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 18)
A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :
1.05 $$\times$$ 10$$-$$34 Js
2.11 $$\times$$ 10$$-$$34 Js
3.16 $$\times$$ 10$$-$$34 Js
4.22 $$\times$$ 10$$-$$34 Js
Explanation
Change in energy
$$ \begin{aligned} &\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{2} \Rightarrow 12.09=\mathrm{E}_{\mathrm{f}}-(-13.6) \Rightarrow \mathrm{E}_{\mathrm{f}}=-1.51 \mathrm{eV} \\\\ &\Rightarrow \frac{-13.6}{n^{2}}=-1.51 \Rightarrow \mathrm{n}^{2}=9 \Rightarrow \mathrm{n}=3 \\\\ &\text { So, } \Delta \mathrm{L}=\mathrm{L}_{\mathrm{f}}-\mathrm{L}_{\mathrm{i}} \\\\ &=\frac{h}{2 \pi}(3-1)=\frac{2 h}{2 \pi}=\frac{h}{\pi}=\frac{6.63 \times 10^{-34}}{3.14}=2.11 \times 10^{-34} \mathrm{Js} \end{aligned} $$
$$ \begin{aligned} &\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{2} \Rightarrow 12.09=\mathrm{E}_{\mathrm{f}}-(-13.6) \Rightarrow \mathrm{E}_{\mathrm{f}}=-1.51 \mathrm{eV} \\\\ &\Rightarrow \frac{-13.6}{n^{2}}=-1.51 \Rightarrow \mathrm{n}^{2}=9 \Rightarrow \mathrm{n}=3 \\\\ &\text { So, } \Delta \mathrm{L}=\mathrm{L}_{\mathrm{f}}-\mathrm{L}_{\mathrm{i}} \\\\ &=\frac{h}{2 \pi}(3-1)=\frac{2 h}{2 \pi}=\frac{h}{\pi}=\frac{6.63 \times 10^{-34}}{3.14}=2.11 \times 10^{-34} \mathrm{Js} \end{aligned} $$
Comments (0)
