A metal sheet of silver is exposed to ultraviolet radiation of wavelength `1810A^(0)`. The threshold wavelength of silver is `2640A^(0)` Calculate the maximum energy of emitted electron.

Options

A. 3.15 eV
B. 4.15 eV
C. 5.15 eV
D. 2.15 eV

Answer

To solve the problem of finding the maximum energy of the emitted electron when a silver metal sheet is exposed to ultraviolet radiation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelength of ultraviolet radiation, \( \lambda = 1810 \, \text{Å} = 1810 \times 10^{-10} \, \text{m} \)
- Threshold wavelength of silver, \( \lambda_0 = 2640 \, \text{Å} = 2640 \times 10^{-10} \, \text{m} \)

2. **Use the Einstein Photoelectric Equation:**
The maximum kinetic energy (KE) of the emitted electrons can be calculated using the equation:
\[ KE = h \nu - h \nu_0 \]
where \( \nu \) is the frequency of the incident radiation and \( \nu_0 \) is the threshold frequency.

3. **Relate Frequency to Wavelength:**
The frequency \( \nu \) can be calculated using the formula:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)).

4. **Calculate Frequencies:**
- For the incident radiation:
\[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{1810 \times 10^{-10}} = \frac{3 \times 10^8}{1.81 \times 10^{-7}} \approx 1.66 \times 10^{15} \, \text{Hz} \]
- For the threshold frequency:
\[ \nu_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{2640 \times 10^{-10}} = \frac{3 \times 10^8}{2.64 \times 10^{-7}} \approx 1.14 \times 10^{15} \, \text{Hz} \]

5. **Calculate the Maximum Kinetic Energy:**
Using the Planck's constant \( h = 6.63 \times 10^{-34} \, \text{J s} \):
\[ KE = h \nu - h \nu_0 = h (\nu - \nu_0) \]
\[ KE = 6.63 \times 10^{-34} \left(1.66 \times 10^{15} - 1.14 \times 10^{15}\right) \]
\[ KE = 6.63 \times 10^{-34} \times 0.52 \times 10^{15} = 3.45 \times 10^{-19} \, \text{J} \]

6. **Convert Kinetic Energy to Electron Volts:**
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE \, (\text{in eV}) = \frac{3.45 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.16 \, \text{eV} \]

### Final Answer:
The maximum energy of the emitted electron is approximately **2.16 eV**.