A light of energy 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…
JEE Main 2023 — Physics Modern Physics
2023integerhard
A light of energy 12.75eV 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 πx×10−17eVs. The value of x is ______ (use h=4.14×10–15eVs,c=3×108ms–1)
Official previous-year question
Held on 1 Feb 2023 · Verified 6 Jul 2026.
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Solution
Let the electron jumped to nth excited state.
In the ground state, energy E=−13.6eV
So, using relation 12.75=13.6[121−n21]
⇒0.9375=[1−n21]⇒n2=16
⇒n=16=4
Now, angular momentum, L=2πnh=2π4h=π2h
=π2×4.14×10−15
=π828×10−17eVs
Hence, the value of x=828.
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