If `lambda_(0)` is the threshold wavelength for photoelectric emission. `lambda` wavelength of light falling on the surface on the surface of metal, and `m` mass of electron. Then de Broglie wavelength of emitted electron is :-

Options

A. `[(h(lambdalambda_(0)))/(2mc(lambda_(0)-lambda))]overset(1)(2)`
B. `[(h(lambda_(0)-lambda))/(2mclambdalambda_(0))]overset(1)(2)`
C. `[(h(lambda-lambda_(0)))/(2mclambdalambda_(0))]overset(1)(2)`
D. `[(hlambdalambda_(0))/(2mc)]overset(1)(2)`

Explanation

To find the de Broglie wavelength of the emitted electron when light of wavelength \( \lambda \) falls on a metal surface, we can use the photoelectric effect and de Broglie's wavelength formula. Here’s a step-by-step solution:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of sufficient energy (or frequency) hits a metal surface, it can eject electrons. The energy of the incident light can be expressed using the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the light.

### Step 2: Apply the Photoelectric Equation
According to Einstein's photoelectric equation:
\[ E = \phi + KE \]
where:
- \( \phi \) is the work function (minimum energy required to eject an electron),
- \( KE \) is the kinetic energy of the emitted electron.

The work function can also be expressed in terms of the threshold wavelength \( \lambda_0 \):
\[ \phi = \frac{hc}{\lambda_0} \]
Thus, we can rewrite the photoelectric equation as:
\[ \frac{hc}{\lambda} = \frac{hc}{\lambda_0} + KE \]

### Step 3: Solve for Kinetic Energy
Rearranging the equation gives us the kinetic energy of the emitted electron:
\[ KE = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]
Factoring out \( hc \):
\[ KE = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

### Step 4: Use the Kinetic Energy to Find the Velocity
The kinetic energy of the emitted electron can also be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

### Step 5: Solve for Velocity
From the above equation, we can solve for \( v \):
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} \]

### Step 6: Find the de Broglie Wavelength
The de Broglie wavelength \( \lambda_d \) of the emitted electron is given by:
\[ \lambda_d = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum \( p \) can be expressed as:
\[ p = mv \]
Thus,
\[ \lambda_d = \frac{h}{mv} \]
Substituting \( v \) from Step 5:
\[ \lambda_d = \frac{h}{m} \cdot \frac{1}{\sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)}} \]
This simplifies to:
\[ \lambda_d = \frac{h \sqrt{m}}{\sqrt{2hc} \sqrt{\left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)}} \]

### Final Expression
After simplification, we can express the de Broglie wavelength of the emitted electron as:
\[ \lambda_d = \frac{h \lambda_0 \lambda}{2mc(\lambda_0 - \lambda)^{1/2}} \]