Question

In an experiment on photoelectric effect a student plots stopping potential $V_0$ against reciprocal of the wavelength $\lambda$ of the incident light for two different metal A and B. These are shown in the figure. Looking at the graphs, you can most appropriately say that :

• A. work function of metal $B$ is greater than that of metal $A$
• B. for light of certain wavelength falling on both metals, maximum kinetic energy of electrons emitted from $A$ will be greater than those emitted from $B$.
• C. work function of metal $A$ is greater than that of metal $B$.
• D. Student data is not correct.

Answer 1

Answer: (A), (B)

work function of metal $B$ is greater than that of metal $A$

Answer 2

The relationship between stopping potential ($V_0$), work function ($\phi$), and the reciprocal of wavelength ($1/\lambda$) in the photoelectric effect is given by the equation: $V_0 = \frac{hc}{e}\left(\frac{1}{\lambda}\right) - \frac{\phi}{e}$ This equation is in the form of a straight line ($y = mx + c$), where $y = V_0$, $x = 1/\lambda$, the slope $m = hc/e$, and the y-intercept $c = -\phi/e$.

From the graph, we observe that the lines for metals A and B are parallel, indicating they have the same slope ($hc/e$), which is consistent with the theory.

The y-intercept ($-\phi/e$) is directly related to the work function. A more negative y-intercept corresponds to a larger work function. Alternatively, the threshold wavelength ($\lambda_{th}$) for photoelectric emission is given by $\lambda_{th} = hc/\phi$. When the stopping potential is zero ($V_0=0$), the incident photon energy is just enough to overcome the work function, i.e., $h\nu = \phi$, or $hc/\lambda = \phi$. Thus, $1/\lambda_{th} = \phi/hc$.

Looking at the graph, the x-intercept (where $V_0=0$) represents the threshold value of $1/\lambda$. For metal A, the x-intercept is at a smaller value of $1/\lambda$ compared to metal B. Let's denote these as $(1/\lambda)_{th, A}$ and $(1/\lambda)_{th, B}$. So, $(1/\lambda)_{th, A} < (1/\lambda)_{th, B}$.

Since $\phi = hc(1/\lambda_{th})$, we have: $\phi_A = hc(1/\lambda)_{th, A}$ $\phi_B = hc(1/\lambda)_{th, B}$

As $(1/\lambda)_{th, A} < (1/\lambda)_{th, B}$, it directly implies that $\phi_A < \phi_B$. Therefore, the work function of metal B is greater than that of metal A. This makes option (A) correct.

Option (B) states that for a certain wavelength, the maximum kinetic energy of electrons emitted from A will be greater than those from B. The maximum kinetic energy is given by $K_{max} = h\nu - \phi$. For a given frequency $\nu$ (or wavelength $\lambda$), $h\nu$ is the same. Since $\phi_A < \phi_B$, then $h\nu - \phi_A > h\nu - \phi_B$, meaning $K_{max, A} > K_{max, B}$. This statement is also true.

However, the question asks for the "most appropriately" say. The difference in work functions is a fundamental property of the metals and is directly represented by the threshold behavior shown in the graph. Option (A) describes this fundamental difference. Option (B) describes a consequence of this difference under specific incident light conditions. Therefore, stating the difference in work functions is a more direct and appropriate conclusion from the graph itself.

Option (C) is incorrect because $\phi_A < \phi_B$. Option (D) is incorrect because the graph is consistent with the photoelectric effect theory.