Step-by-step Derivation

1. Einstein's Photoelectric Equation: The fundamental equation for the photoelectric effect relates the maximum kinetic energy ($K_{max}$) of the emitted photoelectrons to the energy of the incident photon ($E$) and the work function ($\\phi$) of the metal surface: $K_{max} = E - \phi$

2. Energy of a Photon: The energy of a photon can be expressed in terms of its frequency ($\nu$) or its wavelength ($\lambda$): $E = h\nu = \frac{hc}{\lambda}$ where $h$ is Planck's constant and $c$ is the speed of light.

3. Substituting for Photon Energy: Replacing $E$ in the photoelectric equation gives: $K_{max} = \frac{hc}{\lambda} - \phi$

4. Relating Stopping Potential ($V_0$) to Kinetic Energy: The stopping potential, $V_0$, is the minimum potential difference required to stop the photoelectrons with the maximum kinetic energy. The work done by this potential is equal to $K_{max}$: $K_{max} = eV_0$ where $e$ is the elementary charge.

5. Final Equation for Stopping Potential: Combining the above equations, we get: $eV_0 = \frac{hc}{\lambda} - \phi$ Dividing by $e$, we obtain the relationship between $V_0$ and $\lambda$: $V_0 = \left(\frac{hc}{e}\right)\frac{1}{\lambda} - \frac{\phi}{e}$

Analysis of the Graphs

Variation of $V_0$ with $1/\lambda$:

The equation $V_0 = \left(\frac{hc}{e}\right)\frac{1}{\lambda} - \frac{\phi}{e}$ is in the form of a linear equation $y = mx + c'$, where:

• $y = V_0$
• $x = 1/\lambda$
• The slope is $m = \frac{hc}{e}$, which is a positive constant.
• The y-intercept is $c' = -\frac{\phi}{e}$, which is a negative constant.

This means that a graph of $V_0$ versus $1/\lambda$ should be a straight line with a positive slope and a negative y-intercept. Graph A correctly depicts this linear relationship. The graph starts from a certain threshold frequency (or $1/\lambda_{th}$) for which $V_0 = 0$. For values of $1/\lambda$ less than this, there is no photoemission. Graph B is incorrect because it shows a line passing through the origin, implying a zero work function ($\phi=0$), which is not true for metals.

Therefore, option A is correct.

Variation of $V_0$ with $\lambda$:

The equation is $V_0 = \frac{hc}{e\lambda} - \frac{\phi}{e}$. This shows a non-linear relationship between $V_0$ and $\lambda$. Specifically, $V_0$ is inversely proportional to $\lambda$, with a negative offset.

• As $\lambda$ increases, $V_0$ decreases.
• The relationship is not linear, so graph D (a straight line) is incorrect.
• Let's analyze the shape of the curve. The equation is of the form $V_0 = \frac{A}{\lambda} - B$, where $A = hc/e$ and $B = \phi/e$ are positive constants. This is a hyperbolic curve.
• There is a threshold wavelength, $\lambda_{th} = hc/\phi$, above which no photoemission occurs. At $\lambda = \lambda_{th}$, $V_0 = 0$.
• For the curve to be physically meaningful, we need $\lambda \le \lambda_{th}$ and $V_0 \ge 0$.
• The second derivative of $V_0$ with respect to $\lambda$ is $\frac{d^2V_0}{d\lambda^2} = \frac{2hc}{e\lambda^3}$, which is positive for $\lambda > 0$. This indicates that the graph is concave up.

Graph C shows a decreasing, non-linear curve that is concave up and intersects the horizontal axis at a threshold wavelength $\lambda_{th}$. This perfectly matches our analysis.

Therefore, option C is correct.

Conclusion

Based on the analysis of Einstein's photoelectric equation, the correct variations of stopping potential $V_0$ with $\lambda$ and $1/\lambda$ are represented by graphs A and C.