Threshold frequency is ` v_0` and incident frequency is ` v gt v_0` and maximum K.E. is K. If frequency is doubled then, maximum K.E. will be ?

Options

A. Unchanged
B. More than double
C. Less than double
D. double

Answer

To solve the problem regarding the maximum kinetic energy of electrons when the frequency of incident light is doubled, we will use Einstein's photoelectric equation:

\[ K.E. = h\nu - \phi \]

where:
- \( K.E. \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the metal, which can be expressed in terms of the threshold frequency \( \nu_0 \) as \( \phi = h\nu_0 \).

### Step-by-Step Solution:

1. **Initial Conditions**:
- Let the initial frequency of the incident light be \( \nu \) (where \( \nu > \nu_0 \)).
- The maximum kinetic energy of the emitted electrons is given by:
\[ K.E. = h\nu - h\nu_0 \]
- This can be simplified to:
\[ K.E. = h(\nu - \nu_0) \quad \text{(Equation 1)} \]

2. **Doubling the Frequency**:
- Now, if the frequency is doubled, the new frequency becomes \( 2\nu \).
- The maximum kinetic energy with the new frequency is:
\[ K.E.' = h(2\nu) - h\nu_0 \]
- This can be simplified to:
\[ K.E.' = 2h\nu - h\nu_0 \quad \text{(Equation 2)} \]

3. **Relating the Two Energies**:
- We can express \( K.E.' \) in terms of \( K.E. \):
\[ K.E.' = 2h\nu - h\nu_0 = 2(h\nu - h\nu_0) + h\nu_0 \]
- Substituting Equation 1 into this gives:
\[ K.E.' = 2K.E. + h\nu_0 \]

4. **Comparing the Energies**:
- Since \( h\nu_0 \) is a positive quantity (the work function), we can conclude that:
\[ K.E.' > 2K.E. \]
- Therefore, the maximum kinetic energy when the frequency is doubled is more than double the initial maximum kinetic energy.

### Final Answer:
The maximum kinetic energy will be more than double when the frequency is doubled.