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Physics-Photoelectric Effect

NEET Physics Photoelectric Effect MCQ Question

Type: MCQ-numerical-Medium-Class 12

A metal has a threshold frequency u0 u_0 of 5×10145 \times 10^{14} Hz. What is the maximum kinetic energy of photoelectrons emitted when light of frequency 7×10147 \times 10^{14} Hz is incident on it? (Planck's constant, h=6.63×10−34h = 6.63 \times 10^{-34} Js)

A

1.32 eV

B

0.83 eV

C

0.55 eV

D

1.10 eV

Correct Answer

Option A

Detailed Explanation

The maximum kinetic energy Kmax=h(u−u0)K_{max} = h( u - u_0). Substituting the values, Kmax=6.63×10−34×(7×1014−5×1014)=1.32×10−19K_{max} = 6.63 \times 10^{-34} \times (7 \times 10^{14} - 5 \times 10^{14}) = 1.32 \times 10^{-19} J = 1.32 eV.

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