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Physics · Atomic Physics

JEE Main 2024 — 27 January, Shift 2 — Question 43

The threshold frequency of a metal with work function 6.63 eV is :

  1. Option A:

    16×1015 Hz16 \times 10^{15} \mathrm{~Hz}

  2. Option B:

    16×1012 Hz16 \times 10^{12} \mathrm{~Hz}

  3. Option C:

    1.6×1012 Hz1.6 \times 10^{12} \mathrm{~Hz}

  4. Option D:

    1.6×1015 Hz1.6 \times 10^{15} \mathrm{~Hz}

    Correct

Answer: D

Step-by-step solution

ϕ0=hv0\phi_{0}=h v_{0}

6.63×1.6×10−19=6.63×10−34v06.63 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} v_{0}

v0=1.6×10−1910−34\mathrm{v}_{0}=\frac{1.6 \times 10^{-19}}{10^{-34}}

v0=1.6×1015 Hz\mathrm{v}_{0}=1.6 \times 10^{15} \mathrm{~Hz}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Physics
Chapter
Atomic Physics
Topic
Photoelectric Effect

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