A Hydrogen-like atom has atomic number Z. Pho | vedclass

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(B) The energy of the photon emitted during the transition from $n=4$ to $n=3$ is given by $E = 13.6 Z^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)

Vedclass · Accepted Answer

$3$

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(B) The energy of the photon emitted during the transition from $n=4$ to $n=3$ is given by $E = 13.6 Z^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \ eV$.$E = 13.6 Z^2 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 Z^2 \left( \frac{16-9}{144} \right) = 13.6 Z^2 \left( \frac{7}{144} \right) \approx 0.661 Z^2 \ eV$.The work function $W$ of the metal is calculated from the threshold wavelength $\lambda_0 = 310 \ nm$ as $W = \frac{hc}{\lambda_0} = \frac{1240 \ eV \cdot nm}{310 \ nm} = 4 \ eV$.According to Einstein's photoelectric equation,$K_{\max} = E - W$.Given $K_{\max} = 1.95 \ eV$,we have $1.95 = 0.661 Z^2 - 4$.$0.661 Z^2 = 5.95$.$Z^2 = \frac{5.95}{0.661} \approx 9$.Therefore,$Z = 3$.

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