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JEE MAIN 2026 · Electromagnetic Waves

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 38

Updated
Single correct+4 / -1Answer matches NTA’s final key

A monochromatic source of light operating at 15 kW emits 2.5×10222.5 \times 10^{22} photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ______. (Take h=6.6×10−34h = 6.6 \times 10^{-34} J.s and c=3×108c = 3 \times 10^8 m/s).

Choose one

Solution

Solution

Energy per photon: E=15×1032.5×1022=6×10−19E = \dfrac{15 \times 10^3}{2.5 \times 10^{22}} = 6 \times 10^{-19} J.

λ=hcE=6.6×10−34×3×1086×10−19=3.3×10−7 m=330 nm.\lambda = \frac{hc}{E} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}} = 3.3 \times 10^{-7} \text{ m} = 330 \text{ nm}.

330 nm is below the visible range (about 400–700 nm), in the ultraviolet.

Answer: Ultraviolet.

Answer checked against the official answer key. The solution is our own working.

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