Step 1: Understanding the Concept:
The photoelectric effect occurs when incident light provides enough energy to electrons to overcome the work function of a metal.
Einstein's photoelectric equation relates the maximum kinetic energy of the emitted photoelectrons to the energy of the incident photons and the work function of the material.
Step 2: Key Formula or Approach:
Einstein's photoelectric equation is:
\[ K_{\text{max}} = E_{\text{incident}} - \Phi \]
where \( K_{\text{max}} \) is the maximum kinetic energy, \( E_{\text{incident}} = h\nu_{\text{incident}} \) is the energy of incident photons, and \( \Phi \) is the work function.
The work function is related to the threshold frequency \( \nu_0 \) by:
\[ \Phi = h\nu_0 \]
Kinetic energy is related to maximum velocity \( v_{\text{max}} \) by:
\[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 \]
Step 3: Detailed Explanation:
Given that the threshold frequency of the metal is \( \nu_0 = \nu \).
The work function of the metal is:
\[ \Phi = h\nu \]
The frequency of the incident radiation is given as \( \nu_{\text{incident}} = 4\nu \).
The energy of the incident radiation is:
\[ E_{\text{incident}} = h(4\nu) = 4h\nu \]
Now, substitute these into Einstein's photoelectric equation:
\[ K_{\text{max}} = 4h\nu - h\nu \]
\[ K_{\text{max}} = 3h\nu \]
We can also express the maximum kinetic energy in terms of the maximum velocity \( v_{\text{max}} \):
\[ \frac{1}{2} m v_{\text{max}}^2 = 3h\nu \]
We need to solve for \( v_{\text{max}} \). Rearrange the equation:
\[ v_{\text{max}}^2 = \frac{2 \times 3h\nu}{m} \]
\[ v_{\text{max}}^2 = \frac{6h\nu}{m} \]
Taking the square root of both sides gives the maximum velocity:
\[ v_{\text{max}} = \sqrt{\frac{6h\nu}{m}} \]
Step 4: Final Answer:
The maximum velocity of the emitted photoelectrons is \( \sqrt{\frac{6h\nu}{m}} \).