Physics Q43 – JEE-MAIN 2026

Home Physics Dual Nature of MatterQuestion #76

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 :

Clue 1 of 3: Photoelectric Effect

Recall the fundamental principle of the photoelectric effect, which relates the energy of incident photons to the kinetic energy of emitted electrons and the work function of the material.

🧠 Full Solution Path

Step 1: Apply Einstein's Photoelectric Equation✦ Active

According to Einstein's photoelectric equation, the maximum kinetic energy ( $K_{m a x}$ ) of emitted photoelectrons is given by $K_{m a x} = \frac{h c}{λ} - ϕ$ , where $h$ is Planck's constant, $c$ is the speed of light, $λ$ is the wavelength of incident light, and $ϕ$ is the work function of the material.

For the first case:

K_{1} = \frac{h c}{λ_{1}} - ϕ (1)

For the second case:

K_{2} = \frac{h c}{λ_{2}} - ϕ (2)

Step 2: Incorporate the Wavelength Relationship○ Expand

We are given that $λ_{1} = 2 λ_{2}$ . Substitute this into equation (1):

K_{1} = \frac{h c}{2 λ_{2}} - ϕ (3)

💡 Teacher's Secret Hint

Ensure correct substitution of the wavelength relationship.

Step 3: Solve for the Work Function○ Expand

From equation (2), we can express $\frac{h c}{λ_{2}}$ as:

\frac{h c}{λ_{2}} = K_{2} + ϕ (4)

Now, substitute equation (4) into equation (3):

K_{1} = \frac{1}{2} (K_{2} + ϕ) - ϕ

Simplify and solve for $ϕ$ :

2 K_{1} = K_{2} + ϕ - 2 ϕ

2 K_{1} = K_{2} - ϕ

ϕ = K_{2} - 2 K_{1}

This matches option 4.

💡 Teacher's Secret Hint

Be careful with algebraic manipulation, especially when combining terms with $ϕ$ .

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Physics Question 43 – JEE-MAIN 2026

Clue 1 of 3: Photoelectric Effect

🧠 Full Solution Path