The photoelectric threshold frequency of a metal is v. When light of frequency 4v is incident on the metal . The maximum kinetic energy of the emitted photoelectrons is

Options

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

Answer

To solve the problem, we will use the photoelectric effect equation, which relates the maximum kinetic energy of emitted photoelectrons to the frequency of the incident light and the threshold frequency of the metal.

### Step-by-step Solution:

1. **Identify the given values**:
- Threshold frequency of the metal, \( v_0 = v \)
- Frequency of the incident light, \( v = 4v_0 = 4v \)

2. **Use the photoelectric equation**:
The maximum kinetic energy (\( KE_{max} \)) of the emitted photoelectrons can be expressed using the equation:
\[ KE_{max} = h \cdot v - \phi \]
where \( \phi \) is the work function of the metal, which can also be expressed in terms of the threshold frequency:
\[ \phi = h \cdot v_0 \]

3. **Substitute the values into the equation**:
- The energy of the incident light can be calculated as:
\[ KE_{max} = h \cdot (4v) - h \cdot v_0 \]

4. **Replace \( v_0 \) with \( v \)**:
Since \( v_0 = v \), we can substitute this into the equation:
\[ KE_{max} = h \cdot (4v) - h \cdot v \]

5. **Simplify the equation**:
\[ KE_{max} = h \cdot (4v - v) = h \cdot (3v) \]

6. **Final expression for maximum kinetic energy**:
Thus, the maximum kinetic energy of the emitted photoelectrons is:
\[ KE_{max} = 3h \cdot v \]

### Conclusion:
The maximum kinetic energy of the emitted photoelectrons when light of frequency \( 4v \) is incident on the metal is \( 3h \cdot v \).

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