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Photons of energy 10.25 eV fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 5.0 eV. What is the stopping voltage required for these electrons?(a) 10 V(b) 4 V(c) 8 V(d) 5 VThis question was addressed to me in an online interview.Asked question is from Particle Nature of Light : The Photon in section Dual Nature of Radiation and Matter of Physics – Class 12

Answer»

Correct option is (d) 5 V

Easy explanation: The required equation is given as:

Stopping voltage = \(\frac {K_{max}}{e}\).

\(\frac {K_{max}}{e} = \frac {5.0 EV}{e}\)

\(\frac {K_{max}}{e}\) = 5.0 V

Therefore, the stopping POTENTIAL = 5 V.



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