Question

The graph shows stopping potential \(V_s\) versus frequency \(f\) of incident light for three different metals \(X_1, X_2,\) and \(X_3\). Choose the metal which will eject photoelectrons with maximum kinetic energy for a given frequency.

• A. \(X_1\)
• B. \(X_2\)
• C. \(X_3\)
• D. can't be predicted

Answer 1

The Correct Option is A

Concept:
According to Einstein’s photoelectric equation, \[ K_{\text{max}} = eV_s = hf - \phi \] where \(K_{\text{max}}\) = maximum kinetic energy of emitted electrons, \(V_s\) = stopping potential, \(h\) = Planck’s constant, \(f\) = frequency of incident light, \(\phi\) = work function of the metal. The threshold frequency \(f_0\) is related to work function as: \[ \phi = hf_0 \] Thus, \[ K_{\text{max}} = h(f - f_0) \] For a fixed incident frequency, the metal with the lowest threshold frequency (or lowest work function) will give the maximum kinetic energy. Step 1: Interpret the graph. In a stopping potential vs frequency graph: \begin{itemize} \item All metals have the same slope \( \frac{h}{e} \). \item The intercept on the frequency axis represents the threshold frequency. \end{itemize} Step 2: Identify the smallest threshold frequency. From the graph, the line corresponding to \(X_1\) intersects the frequency axis first. Thus, \[ f_{0(X_1)}<f_{0(X_2)}<f_{0(X_3)} \] Step 3: Determine which metal gives maximum kinetic energy. Since \[ K_{\text{max}} = h(f - f_0) \] smaller \(f_0\) gives larger \(K_{\text{max}}\). Therefore, \[ X_1 \text{ produces the maximum kinetic energy.} \]