Light of frequency `v_0` strikes on a metal surface and electrons are ejected with kinetic energy E . If the kinetic energy is to be increased to twice of the given value , the wavelength must be changed to `lamda'` . which is equal to

Options

A. `(hc)/((hv_0-E))`
B. `(hc)/((hv_0-2E))`
C. `(hc)/((hv_0+E))`
D. `v_0^2`

Explanation

To solve the problem, we will use the photoelectric effect equation and manipulate it to find the new wavelength when the kinetic energy of the ejected electrons is doubled.

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
The kinetic energy (KE) of the ejected electrons when light of frequency \( v_0 \) strikes the metal surface is given by:
\[ E = h v_0 - \phi \]
where \( \phi \) is the work function of the metal.

2. **Condition for Doubled Kinetic Energy**:
If the kinetic energy is to be increased to twice the initial value, we write:
\[ 2E = 2(h v_0 - \phi) \]

3. **Express the New Kinetic Energy**:
The new kinetic energy when light of frequency \( v' \) strikes the surface can be expressed as:
\[ KE' = h v' - \phi \]
Setting this equal to the doubled kinetic energy:
\[ h v' - \phi = 2(h v_0 - \phi) \]

4. **Expand and Rearrange the Equation**:
Expanding the right-hand side gives:
\[ h v' - \phi = 2h v_0 - 2\phi \]
Rearranging this, we have:
\[ h v' = 2h v_0 - \phi \]
\[ h v' = 2h v_0 - \phi + \phi \]
\[ h v' = 2h v_0 - \phi + \phi \]
\[ h v' = 2h v_0 - \phi \]

5. **Solve for the New Frequency**:
From the equation \( h v' = 2h v_0 - \phi \), we can isolate \( v' \):
\[ v' = 2v_0 - \frac{\phi}{h} \]

6. **Relate Frequency to Wavelength**:
The relationship between frequency and wavelength is given by:
\[ v = \frac{c}{\lambda} \]
Thus, for the new frequency \( v' \):
\[ v' = \frac{c}{\lambda'} \]
Setting the two expressions for \( v' \) equal gives:
\[ \frac{c}{\lambda'} = 2v_0 - \frac{\phi}{h} \]

7. **Express the New Wavelength**:
Rearranging for \( \lambda' \):
\[ \lambda' = \frac{c}{2v_0 - \frac{\phi}{h}} \]

### Final Expression:
Thus, the new wavelength \( \lambda' \) is given by:
\[ \lambda' = \frac{c}{2v_0 - \frac{\phi}{h}} \]