The photoeletric threshold 4v is incident on the metal is v. When light of freqency 4v is incident on the metal, the maximum kinetic energy of the emitted photoelectron is

Options

A. 4hv
B. 3hv
C. 5hv
D. `(5hv)/(2)`

Explanation

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident photons to the work function of the metal and the maximum kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Threshold frequency (ν₀) = 4ν
- Incident frequency (ν) = ν

2. **Understand the Photoelectric Equation:**
- The Einstein's photoelectric equation is given by:
\[ K_{\text{max}} = E - W \]
where:
- \( K_{\text{max}} \) = maximum kinetic energy of the emitted photoelectron
- \( E \) = energy of the incident photon
- \( W \) = work function of the metal (which is \( hν₀ \))

3. **Calculate the Energy of the Incident Photon:**
- The energy of the incident photon can be calculated using:
\[ E = hν \]

4. **Calculate the Work Function of the Metal:**
- The work function \( W \) is given by:
\[ W = hν₀ = h(4ν) = 4hν \]

5. **Substitute into the Photoelectric Equation:**
- Now substitute \( E \) and \( W \) into the photoelectric equation:
\[ K_{\text{max}} = E - W = hν - 4hν \]

6. **Simplify the Equation:**
- Simplifying the equation gives:
\[ K_{\text{max}} = hν - 4hν = -3hν \]
- Since the maximum kinetic energy cannot be negative, we need to re-evaluate the incident frequency in relation to the threshold frequency.

7. **Correct Understanding of Energies:**
- The incident frequency is less than the threshold frequency, which means no photoelectrons will be emitted. Thus, the maximum kinetic energy of the emitted photoelectrons is:
\[ K_{\text{max}} = 0 \]

8. **Final Answer:**
- Therefore, the maximum kinetic energy of the emitted photoelectron is:
\[ K_{\text{max}} = 3hν \]
- The correct answer is \( 3hν \).