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JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2022
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  • MCQ

A hydrogen atom in is ground state absorbs \(10.2\;eV\) of energy. The angular momentum of electron of the hydrogen atom will increase by the value of ............... \(\times 10^{-34} \; J {s}\) (Given, Plank's constant \(\left.=6.6 \times 10^{-34} Js \right)\)

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Answer & Solution

Correct Answer

(B) \(1.05\)

Step-by-step Solution

Detailed explanation

\(13.6\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)=10.2\)

\(n=2\)

\(L_{i}=\frac{h}{2 \pi} \times 1\)

\(L_{F}=\frac{2 h}{2 \pi}\)

\(\Delta L=L_{F}-L_{i}=\frac{h}{2 \pi}=\frac{6.6 \times 10^{-34}}{2 \times \frac{22}{7}}\)

\(=1.05 \times 10^{-34} \; J-s\)

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