$\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_{1}$ and $\lambda_{2}$, respectively. If $\lambda_{1}=2 \lambda_{2}$ then the work function of material is given by :

(D) $\mathrm{K}_{2}-2 \mathrm{~K}_{1}$

(A) $\mathrm{K}_{2}+2 \mathrm{~K}_{1}$

(B) $2 \mathrm{~K}_{2}-\mathrm{K}_{1}$

(C) $\mathrm{K}_{1}-2 \mathrm{~K}_{2}$

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Modern Physics

K_{1}

and

K_{2}

be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength

λ_{1}

and

λ_{2}

, respectively. If

λ_{1} = 2 λ_{2}

then the work function of material is given by :

(A) $K_{2} + 2 K_{1}$
(B) $2 K_{2} - K_{1}$
(C) $K_{1} - 2 K_{2}$
(D) $K_{2} - 2 K_{1}$ ✓ Correct

Answer: (D)

Solution

From the photoelectric equation,

K_{1} = \frac{h c}{λ _{1}} - ϕ

and

K_{2} = \frac{h c}{λ _{2}} - ϕ

. With

λ_{1} = 2 λ_{2}

\frac{h c}{λ _{2}} = \frac{2 h c}{λ _{1}}

, so

K_{2} = \frac{2 h c}{λ _{1}} - ϕ = 2 (\frac{h c}{λ _{1}} - ϕ) + ϕ = 2 K_{1} + ϕ

. Rearranging, the work function is

ϕ = K_{2} - 2 K_{1}

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option

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