Chemistry · Structure of Atom

JEE Main 2026 — 23 January, Evening Shift — Question 60

The work functions of two metals ( $\mathrm{M}_{\mathrm{A}}$ and $\mathrm{M}_{\mathrm{B}}$ ) are in the $1: 2$ ratio. When these metals are exposed to photons of energy 6 eV , the kinetic energy of liberated electrons of $M_{A}: M_{B}$ is in the ratio of 2.642 : 1. The work functions (in eV ) of $\mathrm{M}_{\mathrm{A}}$ and $\mathrm{M}_{\mathrm{B}}$ are respectively.

Option A:
$3.1,6.2$
Option B:
$2.3,4.6$
Correct
Option C:
$1.4,2.8$
Option D:
$1.5,3.0$

Answer: B

Step-by-step solution

Einstein equation, $\mathrm{KE}_{\max} = E - \phi$

For metal A: $\left(\mathrm{KE}_{\max}\right)_1 = 6 - \phi_1$ …..(1)

For metal B: $\left(\mathrm{KE}_{\max}\right)_2 = 6 - \phi_2$ …..(2)

Given: $\phi_2 = 2\phi_1$

Dividing (1) by (2), we have

\frac{\left(\mathrm{KE}_{\max}\right)_1} {\left(\mathrm{KE}_{\max}\right)_2} = \frac{6-\phi_1}{6-\phi_2} = \frac{2.642}{1}

Substitute $\phi_2 = 2\phi_1$ , we have

\frac{2.642}{1} = \frac{6-\phi_1}{6-2\phi_1}

Cross multiplying, we have

2.642(6-2\phi_1)=6-\phi_1

15.852-5.284\phi_1=6-\phi_1

4.284\phi_1=9.852

\phi_1=2.3\,\mathrm{eV}

\phi_2=4.6\,\mathrm{eV}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Evening Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Photoelectric Effect and Planck's Quantum Theory

Identify the incorrect statements: A. ^24 12 Mg represents 24 protons and 12 neutrons. B. Wavelength of radiation of frequency 4.5 × 10^15…

Answer: B

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