What is work function for photoelectric effect with λ₁=2λ₂?

φ=K₂−2K₁. From K₁=(K₂+φ)/2−φ: 2K₁=K₂−φ.

Why is E₁=E₂/2 when λ₁=2λ₂?

E=hc/λ. Doubling wavelength halves photon energy.

Photoelectric λ₁=2λ₂ kinetic energies K₁ K₂ — work function

JEERANKERS

PhysicsModern Physics

QMCQModern Physics

$K_1$ and $K_2$ are maximum KE of photoelectrons for wavelengths $\lambda_1$ and $\lambda_2$. If $\lambda_1=2\lambda_2$, work function is:

A) $K_2+2K_1$ B) $2K_2-K_1$ C) $K_1-2K_2$ D) $K_2-2K_1$

✅ Correct Answer

D) K₂−2K₁

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Solution

Write Einstein's equations

$K_1=\dfrac{hc}{\lambda_1}-\phi=\dfrac{hc}{2\lambda_2}-\phi\quad\cdots(1)$

$K_2=\dfrac{hc}{\lambda_2}-\phi\quad\cdots(2)$

Eliminate hc/λ₂

From $(2)$: $\dfrac{hc}{\lambda_2}=K_2+\phi$

Sub in $(1)$: $K_1=\dfrac{K_2+\phi}{2}-\phi=\dfrac{K_2}{2}-\dfrac{\phi}{2}$

$2K_1=K_2-\phi\Rightarrow\phi=K_2-2K_1$

📘 Einstein's Photoelectric Equation

$K_{max}=hc/\lambda-\phi$. Two equations eliminate $hc/\lambda_2$. The substitution $hc/\lambda_1=hc/(2\lambda_2)=(K_2+\phi)/2$ leads directly to $\phi=K_2-2K_1$.

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Important Concepts

λ₁=2λ₂ Means E₁=E₂/2E=hc/λ. Longer λ → less energy. E₁=hc/(2λ₂)=E₂/2.

Elimination StrategyExpress both KEs in terms of hc/λ₂ and φ. Eliminate hc/λ₂ to get φ.

Result φ=K₂−2K₁2K₁=K₂−φ → φ=K₂−2K₁.

Physical CheckK₁λ₂). φ>0 requires K₂>2K₁.

FAQs

Why E₁=E₂/2 when λ₁=2λ₂?

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E=hc/λ: doubling λ halves E.

Can φ be negative?

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No. Work function must be positive. K₂>2K₁ is required.

Typical work function values?

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Metals: 2−5 eV. Cesium: ~2.1eV. Gold: ~5.1eV.

What if K₁=0?

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φ=K₂−0=K₂. Light of λ₁ is at threshold, λ₂=λ₁/2 gives K₂=φ.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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