When radiation is incident on a photoelectron emitter , the stopping potential is found to be `9 volts` . If `e//m` for the electrons is `1.8 xx 10^(11) C kg^(-1)` the maximum velocity of the ejected electrons is

Options

A. `6xx10^(5)ms^(-1)`
B. `8xx10^5 ms^(-1)`
C. `10^6 ms^(-1)`
D. `1.8xx10^6ms^(-1)`

Answer

To find the maximum velocity of the ejected electrons when radiation is incident on a photoelectron emitter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Stopping Potential**:
The stopping potential (V₀) is the potential difference required to stop the most energetic photoelectrons emitted from the surface. It is related to the kinetic energy of the electrons.

2. **Use the Formula for Kinetic Energy**:
The kinetic energy (KE) of the ejected electrons can be expressed in terms of the stopping potential:
\[ KE = eV₀ \]
where \( e \) is the charge of the electron and \( V₀ \) is the stopping potential.

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

4. **Set the Two Expressions for Kinetic Energy Equal**:
Equating the two expressions for kinetic energy, we have:
\[ eV₀ = \frac{1}{2} mv^2 \]

5. **Rearrange the Equation to Solve for Velocity**:
Rearranging the equation gives:
\[ v^2 = \frac{2eV₀}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{2eV₀}{m}} \]

6. **Substitute the Given Values**:
We know:
- \( V₀ = 9 \, \text{volts} \)
- \( \frac{e}{m} = 1.8 \times 10^{11} \, \text{C kg}^{-1} \)

Therefore, we can express \( e \) in terms of \( \frac{e}{m} \):
\[ e = \frac{e}{m} \cdot m \]
But since we need \( m \) for the calculation, we can directly use the value of \( \frac{e}{m} \) in our formula:
\[ v = \sqrt{2 \cdot \frac{e}{m} \cdot V₀} \]

7. **Calculate the Maximum Velocity**:
Plugging in the values:
\[ v = \sqrt{2 \cdot (1.8 \times 10^{11}) \cdot 9} \]
\[ v = \sqrt{32.4 \times 10^{11}} \]
\[ v = \sqrt{3.24 \times 10^{12}} = 1.8 \times 10^{6} \, \text{m/s} \]

### Final Answer:
The maximum velocity of the ejected electrons is:
\[ v = 1.8 \times 10^{6} \, \text{m/s} \]