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JEE MAIN 2026 · Atomic Structure

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 73

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

Consider two radiations of wavelengths

  1. λ1=2000\lambda_1 = 2000 Å
  2. λ2=6000\lambda_2 = 6000 Å

The ratio of the energies of these two radiations (E1E2)\left(\frac{E_1}{E_2}\right) is ______ (Nearest integer).

Your answer

Solution

Solution

Photon energy is E=hcλE = \frac{hc}{\lambda}, inversely proportional to wavelength:

E1E2=λ2λ1=60002000=3.\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = \frac{6000}{2000} = 3.

Answer: 3.

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

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