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Atoms and Nuclei question

2023 · 1 Feb · Shift 1 · Q65
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Atoms and Nuclei question

2023 · 1 Feb · Shift 1 · Q65

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
A light of energy 12.75 eV12.75 ~\mathrm{eV}12.75 eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is xπ×10−17 eVs\frac{x}{\pi} \times 10^{-17} ~\mathrm{eVs}πx​×10−17 eVs. The value of xxx is ‾\underline{\hspace{2cm}}​ (use h=4.14×10−15 eVs,c=3×108 ms−1h=4.14 \times 10^{-15} ~\mathrm{eVs}, c=3 \times 10^{8} \mathrm{~ms}^{-1}h=4.14×10−15 eVs,c=3×108 ms−1 ).
Numerical answer
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