Question:
The magnetic field associated with a light wave is given, at the origin, by \(B = B_0\,[\sin(3.14 \times 10^{7})ct + \sin(6.28 \times 10^{7})ct]\). If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(Take \(c = 3 \times 10^{8}\) ms\(^{-1}\), \(h = 6.6 \times 10^{-34}\) J-s)
The magnetic field associated with a light wave is given, at the origin, by \(B = B_0\,[\sin(3.14 \times 10^{7})ct + \sin(6.28 \times 10^{7})ct]\). If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(Take \(c = 3 \times 10^{8}\) ms\(^{-1}\), \(h = 6.6 \times 10^{-34}\) J-s)
(Take \(c = 3 \times 10^{8}\) ms\(^{-1}\), \(h = 6.6 \times 10^{-34}\) J-s)
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Find the two angular frequencies, convert to frequencies, and use the higher one in the photoelectric equation K = hf - work function.
Updated On: Nov 5, 2026
- 6.82 eV
- 12.5 eV
- 8.52 eV
- 7.72 eV
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The Correct Option is D
Solution and Explanation
The two components have angular frequencies \(\omega_1 = 3.14 \times 10^{7}c\) and \(\omega_2 = 6.28 \times 10^{7}c\) rad/s.
\(f = \frac{\omega}{2\pi}\): \(f_1 = 0.5 \times 10^{7} \times 3 \times 10^{8} = 1.5 \times 10^{15}\) Hz, \(f_2 = 1 \times 10^{7} \times 3 \times 10^{8} = 3 \times 10^{15}\) Hz.
Maximum kinetic energy comes from the higher frequency: \(hf_2 = 6.6 \times 10^{-34} \times 3 \times 10^{15} = 1.98 \times 10^{-18}\) J \(\approx 12.4\) eV (using 1 eV = \(1.6 \times 10^{-19}\) J).
\(K_{max} = hf_2 - \phi = 12.4 - 4.7 \approx 7.7\) eV, closest to 7.72 eV.
(The lower frequency gives only \(\approx 6.2 - 4.7 = 1.5\) eV, which is smaller.)
Answer: (4) 7.72 eV
\(f = \frac{\omega}{2\pi}\): \(f_1 = 0.5 \times 10^{7} \times 3 \times 10^{8} = 1.5 \times 10^{15}\) Hz, \(f_2 = 1 \times 10^{7} \times 3 \times 10^{8} = 3 \times 10^{15}\) Hz.
Maximum kinetic energy comes from the higher frequency: \(hf_2 = 6.6 \times 10^{-34} \times 3 \times 10^{15} = 1.98 \times 10^{-18}\) J \(\approx 12.4\) eV (using 1 eV = \(1.6 \times 10^{-19}\) J).
\(K_{max} = hf_2 - \phi = 12.4 - 4.7 \approx 7.7\) eV, closest to 7.72 eV.
(The lower frequency gives only \(\approx 6.2 - 4.7 = 1.5\) eV, which is smaller.)
Answer: (4) 7.72 eV
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