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Physics · Atomic Physics

JEE Main 2026 — 2 April, Evening Shift — Question 63

The surface of sodium metal is irradiated with radiation of wavelength x nm . The kinetic energy of ejected electrons is 2.8×10−20 J2.8 \times 10^{-20} \mathrm{~J}. The work function of sodium is 2.3 eV . The value of x is ____\_\_\_\_ ×102 nm\times 10^{2} \mathrm{~nm}. (Nearest integer) (Given : h=6.6×10−34 J s;1eV=1.60×10−19 J\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s} ; 1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}; c=3.0×108 ms−1\mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1} )

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

hcλ=W+K.E.6.6×10−34×3×108x=2.3×1.6×10−19+2.8×10−2019.8×10−26x=3.96×10−19x=5×10−7x=500×10−9 mx=5×102 nm\begin{aligned} \frac{hc}{\lambda} &= W + K.E. \\ \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{x} &= 2.3 \times 1.6 \times 10^{-19} + 2.8 \times 10^{-20} \\ \frac{19.8 \times 10^{-26}}{x} &= 3.96 \times 10^{-19} \\ x &= 5 \times 10^{-7} \\ x &= 500 \times 10^{-9}\,\mathrm{m} \\ x &= 5 \times 10^{2}\,\mathrm{nm} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Evening Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Photoelectric Effect

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