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. (a) `sqrt((2h)/(m)(lamda_(0)-lamda))`
B. (b) `sqrt((2hc)/(m)(lamda_(0)-lamda))`
C. (c) `sqrt((2hc)/(m)((lamda_(0)-lamda)/(lamdalamda_(0)))`
D. (d) `sqrt((2h)/(m)((1)/lamda_(0)-(1)/lamda)`

Answer

To solve the problem of finding the velocity of photoelectrons ejected from a metal surface when light of wavelength \( \lambda \) is incident, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect describes the phenomenon where electrons are ejected from a metal surface when it is exposed to light of sufficient energy.

2. **Identify the Threshold Wavelength**:
The threshold wavelength \( \lambda_0 \) is the maximum wavelength of light that can cause the ejection of electrons. The energy corresponding to this wavelength is given by:
\[ E_0 = \frac{hc}{\lambda_0} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

3. **Calculate the Energy of Incident Light**:
The energy of the incident light with wavelength \( \lambda \) is given by:
\[ E = \frac{hc}{\lambda} \]

4. **Determine the Kinetic Energy of Ejected Electrons**:
The kinetic energy (KE) of the ejected electrons can be calculated by subtracting the threshold energy from the energy of the incident light:
\[ KE = E - E_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]
Simplifying this, we have:
\[ KE = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

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

6. **Set the Two Expressions for Kinetic Energy Equal**:
Equating the two expressions for kinetic energy gives:
\[ \frac{1}{2} mv^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

7. **Solve 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 \( v \):
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} \]

### Final Expression:
Thus, the velocity of the photoelectron ejected from the metal surface is given by:
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} \]