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Physics-Photoelectric Effect

NEET Physics Photoelectric Effect MCQ Question

Type: MCQ-numerical-Hard-Class 12

For a metal with a threshold frequency u0=5×1014 Hz u_0 = 5 \times 10^{14} \text{ Hz}, what is the work function of the metal in electronvolts (eV)? (Use h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s} and 1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J})

A

2.07 eV

B

3.31 eV

C

5.52 eV

D

1.98 eV

Correct Answer

Option A

Detailed Explanation

The work function ϕ\phi is given by ϕ=hu0\phi = h u_0. Substituting the values, ϕ=6.63×10−34×5×1014=3.315×10−19 J\phi = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 3.315 \times 10^{-19} \text{ J}. Converting to eV, ϕ=3.315×10−191.6×10−19=2.07 eV\phi = \frac{3.315 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.07 \text{ eV}.

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