Photoelectric threshold of silver is `lamda=3800A`. Ultraviolet light of `lamda=2600A` is incident of a silver surface. Calculate:
a. the value of work function in joule and in eV.
b. maximum kinetic energy of the emitted photoelectrons.
c. the maximum velocity of the photoelectrons.
(Mass of the electrons`=9.11xx10^(-31)`).

Options

A. `71.89xx10^(8)`
B. `57.89xx10^(8)`
C. `42.93xx10^(8)`
D. `68.26xx10^(8)`

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Given Data:
- Threshold wavelength of silver, \( \lambda_0 = 3800 \, \text{Å} = 3800 \times 10^{-10} \, \text{m} \)
- Incident wavelength of ultraviolet light, \( \lambda = 2600 \, \text{Å} = 2600 \times 10^{-10} \, \text{m} \)
- Mass of electron, \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)
- Charge of electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

### Part (a): Calculate the Work Function in Joules and eV

1. **Calculate the work function (φ)**:
\[ \phi = \frac{hc}{\lambda_0} \]
Substituting the values:
\[ \phi = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{3800 \times 10^{-10} \, \text{m}} \]
\[ \phi = \frac{1.9878 \times 10^{-25}}{3.8 \times 10^{-7}} \approx 5.23 \times 10^{-19} \, \text{J} \]

2. **Convert the work function to electron volts (eV)**:
\[ \phi \, (\text{eV}) = \frac{\phi \, (\text{J})}{e} = \frac{5.23 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \approx 3.27 \, \text{eV} \]

### Part (b): Calculate the Maximum Kinetic Energy of the Emitted Photoelectrons

1. **Calculate the energy of the incident photon (E)**:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2600 \times 10^{-10} \, \text{m}} \approx 7.65 \times 10^{-19} \, \text{J} \]

2. **Convert the energy to electron volts (eV)**:
\[ E \, (\text{eV}) = \frac{7.65 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \approx 4.78 \, \text{eV} \]

3. **Calculate the maximum kinetic energy (KE_max)**:
\[ KE_{\text{max}} = E - \phi \]
Substituting the values:
\[ KE_{\text{max}} = 4.78 \, \text{eV} - 3.27 \, \text{eV} \approx 1.51 \, \text{eV} \]

### Part (c): Calculate the Maximum Velocity of the Photoelectrons

1. **Use the kinetic energy formula**:
\[ KE_{\text{max}} = \frac{1}{2} mv^2 \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times KE_{\text{max}}}{m}} \]
First, convert \( KE_{\text{max}} \) to Joules:
\[ KE_{\text{max}} \, (\text{J}) = 1.51 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{C} \approx 2.42 \times 10^{-19} \, \text{J} \]

2. **Substituting the values**:
\[ v = \sqrt{\frac{2 \times 2.42 \times 10^{-19} \, \text{J}}{9.11 \times 10^{-31} \, \text{kg}}} \]
\[ v = \sqrt{\frac{4.84 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx \sqrt{5.30 \times 10^{11}} \approx 0.727 \times 10^6 \, \text{m/s} \]

### Summary of Results:
- **Work Function (φ)**: \( 5.23 \times 10^{-19} \, \text{J} \) or \( 3.27 \, \text{eV} \)
- **Maximum Kinetic Energy (KE_max)**: \( 1.51 \, \text{eV} \)
- **Maximum Velocity (v)**: \( 0.727 \times 10^6 \, \text{m/s} \)