The photoelectric threshold wavelength of silver is `3250 xx 10^(-10) m`. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength `2536 xx 10^(-10) m` is
`(Given h = 4.14 xx 10^(6) ms^(-1) eVs` and `c = 3 xx 10^(8) ms^(-1))`

Options

A. `=6xx10^(5)ms^(-1)`
B. `=0.6xx10^(6)ms^(-1)`
C. `=61xx10^(3)ms^(-1)`
D. `=0.3xx10^(6)ms^(-1)`

Explanation

To find the velocity of the electron ejected from a silver surface by ultraviolet light, we will use the photoelectric effect principles and the relevant equations.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Threshold wavelength of silver, \( \lambda_0 = 3250 \times 10^{-10} \, \text{m} \)
- Wavelength of incident light, \( \lambda = 2536 \times 10^{-10} \, \text{m} \)
- Planck's constant, \( h = 4.14 \times 10^{-6} \, \text{ms}^{-1} \, \text{eVs} \)
- Speed of light, \( c = 3 \times 10^{8} \, \text{ms}^{-1} \)
- Mass of electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)

2. **Calculate the Energy of the Incident Photon:**
\[ E = \frac{hc}{\lambda} \]
Substitute the values:
\[ E = \frac{(4.14 \times 10^{-6} \, \text{ms}^{-1} \, \text{eVs})(3 \times 10^{8} \, \text{ms}^{-1})}{2536 \times 10^{-10} \, \text{m}} \]

3. **Calculate the Energy of the Threshold Photon:**
\[ E_0 = \frac{hc}{\lambda_0} \]
Substitute the values:
\[ E_0 = \frac{(4.14 \times 10^{-6} \, \text{ms}^{-1} \, \text{eVs})(3 \times 10^{8} \, \text{ms}^{-1})}{3250 \times 10^{-10} \, \text{m}} \]

4. **Calculate the Kinetic Energy of the Ejected Electron:**
The kinetic energy \( KE \) of the ejected electron is given by:
\[ KE = E - E_0 \]

5. **Relate Kinetic Energy to Velocity:**
The kinetic energy of the electron can also be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = E - E_0 \]

6. **Solve for Velocity \( v \):**
Rearranging gives:
\[ v = \sqrt{\frac{2(E - E_0)}{m}} \]

7. **Substitute and Calculate:**
Substitute the calculated values of \( E \) and \( E_0 \) into the equation to find \( v \).

### Final Calculation:
After performing the calculations, we find:
\[ v \approx 0.6 \times 10^{6} \, \text{m/s} \]

### Conclusion:
The velocity of the ejected electron is approximately \( 0.6 \times 10^{6} \, \text{m/s} \).