Q156 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Physics

Photons of frequencies equal to the frequencies of $H_\beta$ and $H_{\infty}$ lines of hydrogen incident on a photosensitive plate, whose threshold frequency is equal to the frequency of $H_\alpha$ line of hydrogen. The ratio of the maximum kinetic energies of the emitted electrons is

  • Correct answer
    A
    7 : 16
  • B
    3 : 4
  • C
    8 : 27
  • D
    5 : 36

Solution

Energy of photons of $$ \begin{aligned} \mathrm{H}_\alpha \text { line }=\Delta E(3 & \rightarrow 2) \\ & =13.6\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{13.6 \times 5}{36}=1.89 \mathrm{eV} \end{aligned} $$ Energy of photons of $\mathrm{H}_\beta$ line $=\Delta E(4 \rightarrow 2)$ $$ =13.6\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{12 \times 13.6}{64}=2.55 \mathrm{eV} $$ Energy of photons of $\mathrm{H}_{\infty}$ line $=\Delta E(\infty \rightarrow 2)$ $$ =\frac{13.6}{4}=3.4 \mathrm{eV} $$ Ratio of kinetic energy of emitted photons $$ =\frac{255-1.89}{3.4-1.89}=\frac{0.66}{1.51} \approx \frac{0.7}{1.6} \approx \frac{7}{16} $$

Hence, the correct answer is option: 'A'.

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