Question:
A hydrogen atom at the ground level absorbs a photon and is excited to \(n=4\) level. The potential energy of the electron in the excited state is
A hydrogen atom at the ground level absorbs a photon and is excited to \(n=4\) level. The potential energy of the electron in the excited state is
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For hydrogen atom:
\[
E_n=\frac{-13.6}{n^2}\;eV
\]
and
\[
U=2E,\qquad K=-E
\]
where \(U\) is potential energy and \(K\) is kinetic energy.
Updated On: Jun 22, 2026
- \(-0.85\;eV\)
- \(+0.85\;eV\)
- \(-1.7\;eV\)
- \(+1.7\;eV\)
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The Correct Option is C
Solution and Explanation
Step 1: Write the total energy formula for hydrogen atom.
The total energy of an electron in the \(n^{th}\) orbit is \[ E_n=\frac{-13.6}{n^2}\;eV \] For \(n=4\), \[ E_4=\frac{-13.6}{4^2} \] \[ E_4=\frac{-13.6}{16} \] \[ E_4=-0.85\;eV \]
Step 2: Relate potential energy and total energy.
For a hydrogen atom, \[ U=2E \] where \(U\) is potential energy.
Therefore, \[ U=2(-0.85) \] \[ U=-1.7\;eV \]
Step 3: Final conclusion.
Hence, the potential energy of the electron in the excited state is \[ \boxed{-1.7\;eV} \]
The total energy of an electron in the \(n^{th}\) orbit is \[ E_n=\frac{-13.6}{n^2}\;eV \] For \(n=4\), \[ E_4=\frac{-13.6}{4^2} \] \[ E_4=\frac{-13.6}{16} \] \[ E_4=-0.85\;eV \]
Step 2: Relate potential energy and total energy.
For a hydrogen atom, \[ U=2E \] where \(U\) is potential energy.
Therefore, \[ U=2(-0.85) \] \[ U=-1.7\;eV \]
Step 3: Final conclusion.
Hence, the potential energy of the electron in the excited state is \[ \boxed{-1.7\;eV} \]
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