Question:
The radiation corresponding to \(3 \to 2\) transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of \(5 \times 10^{-4}\) T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron \(= 9.1 \times 10^{-31}\) kg)
The radiation corresponding to \(3 \to 2\) transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of \(5 \times 10^{-4}\) T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron \(= 9.1 \times 10^{-31}\) kg)
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Photon energy \(= 13.6\left(\frac14 - \frac19\right)\) eV; maximum kinetic energy from \(r = \frac{mv}{eB}\); then \(\phi = E - K_{max}\).
Updated On: Nov 4, 2026
- 1.36 eV
- 1.88 eV
- 0.82 eV
- 0.16 eV
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The Correct Option is C
Solution and Explanation
Photon energy for \(3 \to 2\): \(E = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \times \frac{5}{36} \approx 1.89\ \mathrm{eV}\).
In the magnetic field \(r = \frac{mv}{eB}\), so \(v_{max} = \frac{eBr}{m}\) and \(K_{max} = \frac{e^2B^2r^2}{2m}\).
In eV: \(K_{max} = \frac{eB^2r^2}{2m} = \frac{1.6 \times 10^{-19} \times (5 \times 10^{-4})^2 \times (7 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.08\ \mathrm{eV}\).
Work function \(\phi = E - K_{max} \approx 1.89 - 1.08 \approx 0.82\ \mathrm{eV}\).
Answer: (3) 0.82 eV
In the magnetic field \(r = \frac{mv}{eB}\), so \(v_{max} = \frac{eBr}{m}\) and \(K_{max} = \frac{e^2B^2r^2}{2m}\).
In eV: \(K_{max} = \frac{eB^2r^2}{2m} = \frac{1.6 \times 10^{-19} \times (5 \times 10^{-4})^2 \times (7 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.08\ \mathrm{eV}\).
Work function \(\phi = E - K_{max} \approx 1.89 - 1.08 \approx 0.82\ \mathrm{eV}\).
Answer: (3) 0.82 eV
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