If `lamda_o and lamda` be the threshold wavelength and wavelength of incident light , the velocity of photoelectron ejected from the metal surface is :

Options

A. `sqrt((2hc)/m((lamda_0-lamda)/(lamda " "lamda_0)))`
B. `sqrt((2hc)/m(lamda_O-lamda))`
C. `sqrt((2h)/m(lamda_O-lamda))`
D. `sqrt((2h)/m(1/lamda_o-1/lamda))`

Answer

To find the velocity of the photoelectron ejected from the metal surface when light of wavelength \(\lambda\) is incident on it, we can use the principles of the photoelectric effect.

### Step-by-Step Solution:

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

2. **Threshold Wavelength**:
The threshold wavelength \(\lambda_0\) is the maximum wavelength of light that can still eject electrons from the metal. The energy corresponding to the threshold wavelength is given by:
\[ E_0 = \frac{hc}{\lambda_0} \]

3. **Kinetic Energy of Ejected Electrons**:
The kinetic energy (KE) of the ejected photoelectrons can be expressed as the difference between the energy of the incident photon and the energy required to eject the electron:
\[ KE = E - E_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]

4. **Expressing Kinetic Energy**:
We can factor out \(hc\) from the equation:
\[ KE = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

5. **Relating Kinetic Energy to Velocity**:
The kinetic energy can also be expressed in terms of the mass \(m\) of the electron and its velocity \(v\):
\[ KE = \frac{1}{2} mv^2 \]

6. **Setting the Equations Equal**:
Now we can set the two expressions for kinetic energy equal to each other:
\[ \frac{1}{2} mv^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

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 gives us the velocity:
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} \]

8. **Final Expression**:
We can rewrite the expression for velocity as:
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)} \]

### Final Answer:
The velocity of the photoelectron ejected from the metal surface is:
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right)} \]