Question:medium
The variation of the stopping potential \( V_0 \) with the frequency \( \nu \) of the incident radiation for four metals A, B, C, and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for which metal?
The variation of the stopping potential \( V_0 \) with the frequency \( \nu \) of the incident radiation for four metals A, B, C, and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for which metal?
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The kinetic energy of photoelectrons is related to the stopping potential: \( K.E. = eV_0 \), where \( e \) is the charge of an electron. The metal with the highest stopping potential at a given frequency will have the maximum kinetic energy of photoelectrons.
Updated On: Jan 13, 2026
- \( A \)
- \( B \)
- \( C \)
- \( D \)
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The Correct Option is A
Solution and Explanation
The stopping potential \( V_0 \) is directly proportional to the kinetic energy of the photoelectrons. As the frequency of the incident radiation rises, so does the stopping potential. Consequently, for a fixed frequency, the metal exhibiting the highest stopping potential will yield photoelectrons with the maximum kinetic energy. In the provided graph, metal A demonstrates the highest stopping potential \( V_0 \) at any given frequency. This observation signifies that metal A liberates photoelectrons with the greatest kinetic energy.
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