Q175 / 180 · COMEDK 2024 (Evening Shift) · Physics

A photon emitted during the de-excitation of electron from a state $\mathrm{n}$ to the second excited state in a hydrogen atom, irradiates a metallic electrode of work function $0.5 \mathrm{~eV}$, in a photocell, with a stopping voltage of $0.47 \mathrm{~V}$. Obtain the value of quantum number of the state '$n$'.

  • Correct answer
    A
    5
  • B
    6
  • C
    4
  • D
    3

Solution

The energy of a photon emitted during the transition from state $n$ to the second excited state ($n_f = 3$) in a hydrogen atom is given by the Rydberg formula:

$E = 13.6 \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right) \text{ eV}$

$E = 13.6 \left( \dfrac{1}{3^2} - \dfrac{1}{n^2} \right) = 13.6 \left( \dfrac{1}{9} - \dfrac{1}{n^2} \right) \text{ eV}$

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is $K_{max} = E - \Phi$, where $\Phi$ is the work function. Given $K_{max} = e V_s = 0.47 \text{ eV}$ and $\Phi = 0.5 \text{ eV}$:

$0.47 = E - 0.5$

$E = 0.47 + 0.5 = 0.97 \text{ eV}$

Equating the two expressions for $E$:

$13.6 \left( \dfrac{1}{9} - \dfrac{1}{n^2} \right) = 0.97$

$\dfrac{1}{9} - \dfrac{1}{n^2} = \dfrac{0.97}{13.6} \approx 0.07132$

$\dfrac{1}{n^2} = \dfrac{1}{9} - 0.07132 \approx 0.11111 - 0.07132 = 0.03979$

$n^2 = \dfrac{1}{0.03979} \approx 25.13$

$n \approx 5$

Answer: 5

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

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