Verified Solution • 2025–26 Session moderate

QJEEM22Ph99c8: JEE Main 2022 Physics - Class 12 JEEM22Ph99c8 Solution: A hydrogen atom in ground state absorbs 12.09 eV of energy. The …

A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :

Multiple Choice Options

  • A 1.05 ×\times 10 −- 34 Js
  • B 2.11 ×\times 10 −- 34 Js
  • C 3.16 ×\times 10 −- 34 Js
  • D 4.22 ×\times 10 −- 34 Js

Step-by-Step Verified Solution

Change in energy

amp;ΔE=Ef−E2⇒12.09=Ef−(−13.6)⇒Ef=−1.51eVamp;⇒−13.6n2=−1.51⇒n2=9⇒n=3amp; So, ΔL=Lf−Liamp;=h2π(3−1)=2h2π=hπ=6.63×10−343.14=2.11×10−34Js\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}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring