JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Atomic Structure

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.

Answer: (B)

Solution

With photon energy

6

eV and work functions satisfying

ϕ_{B} = 2 ϕ_{A}

, the photoelectron energies are

6 - ϕ_{A}

and

6 - 2 ϕ_{A}

. Setting their ratio to

2.642

\frac{6 - ϕ _{A}}{6 - 2 ϕ _{A}} = 2.642

, gives

4.284 ϕ_{A} = 9.852

, so

ϕ_{A} = 2.3

eV and

ϕ_{B} = 4.6

eV.

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option