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(C) Given,stopping potential $V_{0} = 10 \ V$ and work function $W = 2.75 \ eV$.According to Einstein's photoelectric equation,the energy of the incid

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$4$

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(C) Given,stopping potential $V_{0} = 10 \ V$ and work function $W = 2.75 \ eV$.According to Einstein's photoelectric equation,the energy of the incident photon is $E = h
u = eV_{0} + W$.Substituting the values,$E = 10 \ eV + 2.75 \ eV = 12.75 \ eV$ ..... $(i)$.For a hydrogen atom,the energy of a photon emitted during a transition from state $n$ to the ground state $(n=1)$ is given by $h
u = E_{n} - E_{1}$.Since $E_{n} = -\frac{13.6}{n^{2}} \ eV$,we have $h
u = -\frac{13.6}{n^{2}} - (-13.6) = 13.6 \left(1 - \frac{1}{n^{2}}ight) \ eV$.Equating this to the energy from $(i)$,$13.6 \left(1 - \frac{1}{n^{2}}ight) = 12.75$.$1 - \frac{1}{n^{2}} = \frac{12.75}{13.6} = 0.9375$.$\frac{1}{n^{2}} = 1 - 0.9375 = 0.0625$.$n^{2} = \frac{1}{0.0625} = 16$.Therefore,$n = 4$.

An electron in the hydrogen atom jumps from excited state $n$ to the ground state. The wavelength so emitted illuminates a photosensitive material having work function $2.75 \ eV$. If the stopping potential of the photoelectron is $10 \ V$,then the value of $n$ is

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