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Physics · Atomic Physics

JEE Main 2026 — 24 January, Evening Shift — Question 33

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V . If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V . The wavelength of first light is ____\_\_\_\_ m. ( h=6.63×10−34 J.s,e=1.6×10−19C,c=3×108 m/s\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s} )

  1. Option A:

    2.9×10−82.9 \times 10^{-8}

  2. Option B:

    2.2×10−82.2 \times 10^{-8}

  3. Option C:

    3.1×10−73.1 \times 10^{-7}

  4. Option D:

    2.5×10−72.5 \times 10^{-7}

    Correct

Answer: D

Step-by-step solution

q.(3.2)=hcλ−ϕ\mathrm{q} .(3.2)=\frac{\mathrm{hc}}{\lambda}-\phi

\mathrm{q}(0.7)=\frac{\mathrm{hc}}{2 \lambda}-\phi \end{gathered}$$ Eq. (1) - Eq. (2) $\mathrm{q} \cdot(2.5)=\frac{\mathrm{hc}}{2 \lambda}$ $2.5=\left(\frac{\mathrm{hc}}{\mathrm{e}}\right)\left(\frac{1}{2 \lambda}\right)$

\begin{aligned} & 2.5=\frac{12400}{2(\lambda)} & \lambda=\frac{12400}{5} \AA & \lambda=2480 \AA & \lambda=2.48 \times 10^{-7} \mathrm{~m} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Evening Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Photoelectric Effect
When a light of a given wavelength falls on a metallic surface the… | JEE Main 2026 PYQ with Solution · DhiX AI