Consider two radiations of wavelengths : λ 1 =2000 AA λ 2 =6000 AA The ratio of the energies of these two radiations (frac E 1 E 2 ) is \ \ \ \ . (Nearest integer)

Chemistry · Structure of Atom

JEE Main 2026 — 8 April, Evening Shift — Question 61

Consider two radiations of wavelengths :

$\lambda_{1}=2000 \AA$

$\lambda_{2}=6000 \AA$

The ratio of the energies of these two radiations $\left(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\right)$ is $\_\_\_\_$ . (Nearest integer)

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{E}_{\mathrm{photon}}=\frac{\mathrm{hc}}{\lambda}$ $\Rightarrow \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$ $\Rightarrow \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{6000}{2000}$ $\Rightarrow \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=3$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Electromagnetic Waves and Spectra