Step 1: Understanding the Concept:
Einstein's photoelectric equation relates the energy of incident photons to the work function of the material and the maximum kinetic energy of the emitted electrons.
The energy of a photon is inversely proportional to its wavelength.
When the wavelength decreases, the photon energy increases, resulting in higher maximum kinetic energy and higher speed of electrons.
Step 2: Key Formula or Approach:
Photoelectric equation: \[\frac{hc}{\lambda} = \phi + K_{max} = \phi + \frac{1}{2}mv^2\]
where \(\phi\) is the work function.
Step 3: Detailed Explanation:
Case 1: \[\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi \quad \dots (1)\]
Case 2: Wavelength becomes \(\lambda' = \frac{3\lambda}{4}\). The new speed is \(v'\).
\[\frac{1}{2}m(v')^2 = \frac{hc}{3\lambda/4} - \phi = \frac{4}{3} \left(\frac{hc}{\lambda}\right) - \phi \quad \dots (2)\]
From (1), \(\frac{hc}{\lambda} = \frac{1}{2}mv^2 + \phi\). Substitute this into (2):
\[\frac{1}{2}m(v')^2 = \frac{4}{3} \left(\frac{1}{2}mv^2 + \phi\right) - \phi\]
\[\frac{1}{2}m(v')^2 = \frac{4}{3} \left(\frac{1}{2}mv^2\right) + \frac{4}{3}\phi - \phi\]
\[\frac{1}{2}m(v')^2 = \frac{4}{3} \left(\frac{1}{2}mv^2\right) + \frac{1}{3}\phi\]
Dividing by \(\frac{1}{2}m\):
\[(v')^2 = \frac{4}{3}v^2 + \frac{2\phi}{3m}\]
Since \(\phi>0\), it follows that:
\[(v')^2>\frac{4}{3}v^2\]
\[v'>\sqrt{\frac{4}{3}}v\]
Therefore, the new speed is greater than \(v\sqrt{4/3}\). This matches option (A).
Step 4: Final Answer:
Due to the presence of the work function, the increase in kinetic energy is more than proportional to the increase in photon energy, making the final speed \(v'>v\sqrt{4/3}\).