CBSE Class 12 Physics 2024: question 22

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Step-by-step solution

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1. (a) Einstein's photoelectric equation:
   $$K_{max}=h\nu -{\varphi }_0=h\nu -h{\nu }_0,\text{with }{K}_{max}=e{V}_0$$

   ½ mark

2. Millikan's test: Millikan measured the stopping potential $V_0$ for different frequencies $\nu$ using metals such as sodium.

   The equation predicts $V_0=\frac{h}{e}\nu -\frac{{\varphi }_0}{e}$. Millikan's $V_0$ versus $\nu$ graph was indeed a straight line.

   Its slope $h/e$ gave a value of $h$ that agreed closely with Planck's constant. This verified the equation.

   1 mark

3. (b) $K_{max}$ cannot be negative, so $h\nu -{\varphi }_0\ge 0$, which gives $\nu \ge \frac{{\varphi }_0}{h}={\nu }_0$.

   A photon gives all its energy to a single electron. If $h\nu <{\varphi }_0$, that energy is not enough to remove the electron, however intense the light.

   Hence emission occurs only above a minimum frequency ${\nu }_0$, which is the threshold frequency of the surface.

   1½ marks

Answer: (a) $K_{max}=h\nu -{\varphi }_0$ (= eV₀), verified by Millikan's linear V₀–ν plot whose slope h/e gave Planck's constant; (b) K_max ≥ 0 requires ν ≥ ν₀ = φ₀/h

Revise the chapter

• Chapter: Dual Nature of Radiation and Matter (notes, formulas, where marks are lost and every past question)
• Topic: Einstein's Photoelectric Equation