Radiation with energy equal to the separation energy of an electron in $H -$ atom in first excited state, is incident on a metal electrode. Work function of the metal is equal to the separation energy of an electron in $H e^{+}$ ion in fourth excited state. Find the stopping potential (in V) for such arrangement ?
(Answer upto one decimal point)

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Solution

Separation energy is nothing but the energy of the electron in any excited state.

For, $H -$ atom first excited state $(n = 2), E_{s_{1}} = 3.4 eV$
For $H e^{+}$ ion, fourth excited state $(n = 5), E_{s_{2}} = 13.6 \times \frac{2^{2}}{5^{2}}$
$= 2.17 eV$
where, $E_{s}$ is separation energy
Now, from photoelectric equation,
$K . E ._{max} = E_{i} - E_{0}$
where, $E_{i} -$ incident energy
$E_{0} -$ Threshold energy (work function)
So, stopping potential $= \frac{K . E ._{max}}{e} = (3.4 - 2.17) V$
$≅ 1.2 V$

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