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(C) The energy of an electron in the $n^{	ext{th}}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$

Vedclass · Accepted Answer

$6$

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(C) The energy of an electron in the $n^{	ext{th}}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$. For the second orbit $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$. The energy of the emitted photon is the difference between these two states: $E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$. According to Einstein's photoelectric equation,$h
u = \phi_0 + eV_s$,where $h
u$ is the photon energy,$\phi_0$ is the work function,and $V_s$ is the stopping potential. Given $\phi_0 = 4.2 \ eV$ and $h
u = 10.2 \ eV$,we have $10.2 \ eV = 4.2 \ eV + eV_s$. $eV_s = 10.2 \ eV - 4.2 \ eV = 6 \ eV$. Therefore,the stopping potential $V_s = 6 \ V$.

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