The graph shows variation of stopping potential \(V_0\) with the frequency \(\nu\) of the incident radiation for three photosensitive metals \(X_1\), \(X_2\) and \(X_3\). Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?
All the metals will give out photo electrons with same kinetic energies.
Show Solution
The Correct Option isA
Solution and Explanation
Step 1: Understanding the Concept:
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons depends on the energy of the incident photons and the work function of the metal.
The stopping potential graph intercepts the frequency axis at the threshold frequency (\(\nu_0\)), which is directly proportional to the work function. Step 2: Key Formula or Approach:
Einstein's photoelectric equation is:
\(K_{\max} = h\nu - \Phi\)
where \(\Phi = h\nu_0\) is the work function of the metal, and \(\nu_0\) is the threshold frequency (the x-intercept on the \(V_0\) vs \(\nu\) graph). Step 3: Detailed Explanation:
For the same wavelength of incident radiation, the frequency \(\nu = \frac{c}{\lambda}\) is constant for all three metals.
Therefore, the term \(h\nu\) is identical for \(X_1, X_2\), and \(X_3\).
To maximize the kinetic energy \(K_{\max}\), the work function \(\Phi\) must be as small as possible.
From the given graph, the x-intercepts (threshold frequencies) follow the order:
\(\nu_{01}<\nu_{02}<\nu_{03}\).
Since \(\Phi = h\nu_0\), the work functions follow the same order:
\(\Phi_{X_1}<\Phi_{X_2}<\Phi_{X_3}\).
Because metal \(X_1\) has the lowest work function, less energy is required to eject the electron, leaving more energy as the kinetic energy of the emitted electron. Step 4: Final Answer:
Metal \(X_1\) will emit photoelectrons with the greatest maximum kinetic energy.
