The kinetic energy of an electron is E when the incident wavelength is `lamda` To increase ti KE of the electron to 2E, the incident wavelength must be

Options

A. `2lambda`
B. `lambda//2`
C. `((hclambda))/((Elambda+hc))`
D. `((hclambda))/((Elambda-hc))`

Answer

To solve the problem of finding the new wavelength required to increase the kinetic energy of an electron from E to 2E, we can use the principles of the photoelectric effect. Here's a step-by-step solution:

### Step 1: Understand the Photoelectric Equation
The photoelectric equation relates the energy of the incident photon to the work function and the kinetic energy of the emitted electron:
\[ E_k = h\nu - \phi_0 \]
where:
- \( E_k \) is the kinetic energy of the electron,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi_0 \) is the work function of the material.

### Step 2: Relate Frequency and Wavelength
The frequency \( \nu \) can be expressed in terms of wavelength \( \lambda \):
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Thus, we can rewrite the kinetic energy equation as:
\[ E_k = \frac{hc}{\lambda} - \phi_0 \]

### Step 3: Set Up the Equations
For the first case, when the kinetic energy is \( E \) and the wavelength is \( \lambda \):
\[ E = \frac{hc}{\lambda} - \phi_0 \]
This can be rearranged to:
\[ \frac{hc}{\lambda} = E + \phi_0 \]
Let this be Equation (1).

For the second case, when the kinetic energy is \( 2E \) and the new wavelength is \( \lambda' \):
\[ 2E = \frac{hc}{\lambda'} - \phi_0 \]
Rearranging gives:
\[ \frac{hc}{\lambda'} = 2E + \phi_0 \]
Let this be Equation (2).

### Step 4: Subtract the Two Equations
Now, we can subtract Equation (1) from Equation (2):
\[ \frac{hc}{\lambda'} - \frac{hc}{\lambda} = (2E + \phi_0) - (E + \phi_0) \]
This simplifies to:
\[ \frac{hc}{\lambda'} - \frac{hc}{\lambda} = E \]
Factoring out \( hc \) gives:
\[ hc \left( \frac{1}{\lambda'} - \frac{1}{\lambda} \right) = E \]

### Step 5: Solve for the New Wavelength
Rearranging the above equation, we find:
\[ \frac{1}{\lambda'} = \frac{1}{\lambda} + \frac{E}{hc} \]
Taking the reciprocal gives:
\[ \lambda' = \frac{hc \lambda}{E \lambda + hc} \]

### Conclusion
Thus, the new wavelength \( \lambda' \) required to increase the kinetic energy of the electron to \( 2E \) is:
\[ \lambda' = \frac{hc \lambda}{E \lambda + hc} \]