CBSE Class 12 Physics: Photoelectric Effect Marking Scheme

Governing Law: Einstein’s Photoelectric Equation is given by:

Where $K_{max}$ is the maximum kinetic energy, $h$ is Planck’s constant, $\nu$ is the frequency of incident light, and $\Phi_0$ is the work function of the metal.

(a) Threshold Frequency Observation: [1 Mark]

• Step 1: For photoelectric emission to occur, $K_{max}$ must be $\ge 0$.
• Step 2: Substituting into the equation: $h\nu - \Phi_0 \ge 0 \implies h\nu \ge \Phi_0$.
• Step 3: Thus, $\nu \ge \frac{\Phi_0}{h}$. The term $\frac{\Phi_0}{h}$ is defined as the threshold frequency ($\nu_0$). If $\nu < \nu_0$, the energy of the incident photon is insufficient to overcome the work function, hence no emission occurs.

(b) Frequency vs. Intensity: [1 Mark]

• Step 1: From $K_{max} = h\nu - \Phi_0$, it is evident that $K_{max}$ depends linearly on the frequency ($\nu$) of the incident radiation.
• Step 2: Intensity of light is proportional to the number of photons incident per unit area per unit time. Increasing intensity increases the number of photoelectrons emitted (photoelectric current), but does not change the energy of individual photons.
• Step 3: Therefore, $K_{max}$ is independent of intensity and depends solely on the frequency of the incident light.

(c) Cut-off Voltage ($V_0$) vs. Frequency ($\nu$) Curve: [1 Mark]

• Step 1: We know $K_{max} = eV_0$, where $V_0$ is the stopping potential.
• Step 2: Substituting this into Einstein's equation: $eV_0 = h\nu - \Phi_0$.
• Step 3: Rearranging for $V_0$: $$V_0 = \left(\frac{h}{e}\right)\nu - \frac{\Phi_0}{e}$$ Comparing this with the equation of a straight line ($y = mx + c$), the slope of the $V_0$ vs. $\nu$ graph is $\frac{h}{e}$, which is a universal constant.