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?
- \(X_1\)
- \(X_2\)
- \(X_3\)
- All the metals will give out photo electrons with same kinetic energies.
The Correct Option is A
Solution and Explanation
Concept:
From photoelectric effect, \[ K_{\max} = eV_0 = h\nu - \phi \] where \( \phi \) is the work function.
For the same frequency (or wavelength), \(h\nu\) is constant. So, \[ K_{\max} \propto (h\nu - \phi) \] Thus, smaller work function \( \phi \) ⇒ greater kinetic energy.
Graph Analysis:
In the graph of \(V_0\) vs \( \nu \):
- The slope is same for all (since slope = \( \frac{h}{e} \))
- The x-intercept gives threshold frequency \( \nu_0 = \frac{\phi}{h} \)
So, the metal with smallest threshold frequency has smallest work function.
Conclusion:
The line which cuts the frequency axis at the leftmost point corresponds to the smallest \( \phi \). Hence, that metal gives maximum kinetic energy.
From the graph, this corresponds to:
\(X_1\)
Final Answer:
\[ \boxed{X_1} \]
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