If `lambda_(0)` is the threshold wavelnegth of a metal and `lambda` is the wavelength of the incident radiation, the maximum velocity of the ejected electron from the metal would be

Options

A. `[ ( 2hc)/( m )((lambda_(0) - lambda)/( lambda_(0) lambda)) ]^(1//2)`
B. `[ ( 2hc)/( m )((lambda-lambda_(0) )/( lambdalambda_(0))) ]^(1//2`
C. `[ ( 2hc)/( m ) ( lambda - lambda_(0) ) ]^(1//2)`
D. `[ ( 2hc)/( m ) ( lambda_(0) - lambda) ]^(1//2)`

Answer

To solve the problem of finding the maximum velocity of the ejected electron from a metal when given the threshold wavelength (λ₀) and the wavelength of the incident radiation (λ), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Threshold Wavelength**:
- The threshold wavelength (λ₀) is the minimum wavelength of light required to eject an electron from the metal surface. If the wavelength of the incident radiation (λ) is greater than λ₀, no electrons will be ejected.

2. **Photon Energy**:
- The energy of a photon can be expressed using the equation:
\[ E = \frac{hc}{\lambda} \]
- Where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the incident radiation.

3. **Kinetic Energy of Ejected Electrons**:
- The kinetic energy (KE) of the ejected electrons can be described by the equation:
\[ KE = E - \phi \]
- Where \(\phi\) (the work function) is the energy required to remove an electron from the metal, which can be expressed as:
\[ \phi = \frac{hc}{\lambda_0} \]

4. **Substituting the Energy Values**:
- Therefore, the kinetic energy of the ejected electron can be rewritten as:
\[ KE = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]

5. **Expressing Kinetic Energy in Terms of Velocity**:
- The kinetic energy can also be expressed in terms of the velocity (v) of the ejected electron:
\[ KE = \frac{1}{2} mv^2 \]
- Where \(m\) is the mass of the electron.

6. **Equating the Two Expressions for Kinetic Energy**:
- Setting the two expressions for kinetic energy equal gives us:
\[ \frac{1}{2} mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]

7. **Solving for Velocity**:
- Rearranging the equation to solve for \(v^2\):
\[ v^2 = \frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]
- Taking the square root to find \(v\):
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} \]

8. **Final Expression**:
- This can also be expressed as:
\[ v = \sqrt{\frac{2hc}{m} \cdot \frac{\lambda_0 - \lambda}{\lambda \cdot \lambda_0}} \]
- Thus, the maximum velocity of the ejected electron is:
\[ v = \sqrt{\frac{2hc(\lambda_0 - \lambda)}{m \lambda \lambda_0}} \]

### Conclusion:
The maximum velocity of the ejected electron is given by:
\[ v = \sqrt{\frac{2hc(\lambda_0 - \lambda)}{m \lambda \lambda_0}} \]