# A photon emitted during the de-excitation of electron from a state $n$ to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function $2\; eV$, in a photo cell, with a stopping potential of $0.55\; V$. Obtain the value of the quantum number of the state $n$.

From einstein photoelectric equation, We know that

$hv=\phi _{o}+eV_{s}$

Where, $\phi$ is the work function

$eV_{s}$ is stopping potential

So,       $h\nu=2+0.55$

$h\nu=0.55\; eV$

We have, $E_{n}=\frac{-13.6}{n^{2}}$

Then, the energy difference, $\Delta E=-3.4-(-2.55)eV=-0.85\; eV$

Therefore,         $\frac{-13.6}{n^{2}}=0.85$

$\therefore \; \; n=4$

The value of the quantum number of the state $n$ is $4$.

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