CBSE Class 12 Physics 2025: question 26

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Step-by-step solution

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1. (a) Work function (${\varphi }_0$): the minimum energy that must be supplied to an electron to just release it from the surface of the metal.

   ½ mark

2. From Einstein's equation, $e{V}_0=h\nu -{\varphi }_0\Rightarrow V_0=\frac{h}{e}\nu -\frac{{\varphi }_0}{e}$. The $V_0$–$\nu$ graph is a straight line. Extended back, it cuts the $V_0$ axis at $-{\varphi }_0/e$, so ${\varphi }_0=e\times |\text{intercept on }{V}_0\text{ axis}|$. Equivalently, the intercept on the $\nu$ axis is the threshold frequency ${\nu }_0$, and ${\varphi }_0=h{\nu }_0$.

   1 mark

3. (b) Energy of the incident photon: $\frac{hc}{\lambda }={\varphi }_0+e{V}_0=2.4+0.6=3.0$ eV.

   ½ mark

4. $\lambda =\frac{hc}{E}=\frac{6.63\times 10^{-34}\times 3\times 10^8}{3.0\times 1.6\times 10^{-19}}$

   ½ mark

5. $\lambda \approx 4.14\times 10^{-7}$ m $\approx 414$ nm.

   ½ mark

Answer: (a) φ₀ = e × |V₀-intercept| (or hν₀ from the ν-intercept); (b) λ ≈ 414 nm (≈ 4.14 × 10⁻⁷ m).

Revise the chapter

• Chapter: Dual Nature of Radiation and Matter (notes, formulas, where marks are lost and every past question)