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(D) The energy of an electron in the $n^{th}$ state of a hydrogen atom is given by $E_n = \frac{-13.6}{n^2} \, eV$.The energy released when the electr

Vedclass · Accepted Answer

$7.0$

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(D) The energy of an electron in the $n^{th}$ state of a hydrogen atom is given by $E_n = \frac{-13.6}{n^2} \, eV$.The energy released when the electron jumps from $n = 3$ to $n = 1$ is:$E = E_3 - E_1 = \left( \frac{-13.6}{3^2} \right) - \left( \frac{-13.6}{1^2} \right) = -1.51 + 13.6 = 12.09 \, eV \approx 12.1 \, eV$.According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted photoelectrons is $K_{max} = E - \phi_0$,where $\phi_0$ is the work function.Given $\phi_0 = 5.1 \, eV$,we have:$K_{max} = 12.1 \, eV - 5.1 \, eV = 7.0 \, eV$.The stopping potential $V_0$ is related to the maximum kinetic energy by $K_{max} = eV_0$.Therefore,$V_0 = 7.0 \, V$.

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