JEE Main 2026 — Jan 23, Shift 2 — Chemistry solution

JEE Main2026Jan 23, Shift 2Chemistry

The work functions of two metals ( $M_{A}$ and $M_{B}$ ) 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 $M_{A}$ and $M_{B}$ are respectively.

A
3.1, 6.2
B
2.3, 4.6
C
1.4, 2.8
D
1.5, 3.0

Solution

From the photoelectric equation: $K E_{ma x} = E - ϕ$ .

Let $ϕ_{A} = ϕ_{1}$ and $ϕ_{B} = ϕ_{2} = 2 ϕ_{1}$ .

$\frac{( K E _{ma x} ) _{A}}{( K E _{ma x} ) _{B}} = \frac{6 - ϕ _{1}}{6 - 2 ϕ _{1}} = \frac{2.642}{1}$

$6 - ϕ_{1} = 2.642 (6 - 2 ϕ_{1})$

$6 - ϕ_{1} = 15.852 - 5.284 ϕ_{1}$

$4.284 ϕ_{1} = 9.852$

$ϕ_{1} \approx 2.3$ eV, $ϕ_{2} \approx 4.6$ eV.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →