The maximum wavelength of a beam of light can be used to produce photo electric effect on a metal is 250 nm. The energy of the electrons in Joule emitted from the surface of the metal when a beam of light of wavelength 200 nm is used `[h=6.62xx10^(-34)Js,C=3xx10^(8)ms^(-1)]`

Options

A. `89.61xx10^(-22)`
B. `69.81xx10^(-22)`
C. `18.96xx10^(-20)`
D. `19.86xx10^(-20)`

Answer

To solve the problem, we need to calculate the energy of the emitted electrons when light of wavelength 200 nm is used on a metal that has a maximum wavelength of 250 nm for the photoelectric effect.

### Step-by-step Solution:

1. **Identify the given values:**
- Maximum wavelength for photoelectric effect, \( \lambda_0 = 250 \, \text{nm} = 250 \times 10^{-9} \, \text{m} \)
- Wavelength of incident light, \( \lambda = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m} \)
- Planck's constant, \( h = 6.62 \times 10^{-34} \, \text{Js} \)
- Speed of light, \( c = 3 \times 10^{8} \, \text{ms}^{-1} \)

2. **Calculate the energy of the incident photons:**
The energy of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
Substituting the values for \( \lambda = 200 \times 10^{-9} \, \text{m} \):
\[ E = \frac{(6.62 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{ms}^{-1})}{200 \times 10^{-9} \, \text{m}} \]
\[ E = \frac{1.986 \times 10^{-25} \, \text{Jm}}{200 \times 10^{-9} \, \text{m}} \]
\[ E = 9.93 \times 10^{-19} \, \text{J} \]

3. **Calculate the work function (Φ) of the metal:**
The work function can be calculated using the maximum wavelength:
\[ E_0 = \frac{hc}{\lambda_0} \]
Substituting \( \lambda_0 = 250 \times 10^{-9} \, \text{m} \):
\[ E_0 = \frac{(6.62 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{ms}^{-1})}{250 \times 10^{-9} \, \text{m}} \]
\[ E_0 = \frac{1.986 \times 10^{-25} \, \text{Jm}}{250 \times 10^{-9} \, \text{m}} \]
\[ E_0 = 7.944 \times 10^{-19} \, \text{J} \]

4. **Calculate the kinetic energy (KE) of the emitted electrons:**
The kinetic energy of the emitted electrons is given by:
\[ KE = E - E_0 \]
Substituting the values of \( E \) and \( E_0 \):
\[ KE = 9.93 \times 10^{-19} \, \text{J} - 7.944 \times 10^{-19} \, \text{J} \]
\[ KE = 1.986 \times 10^{-19} \, \text{J} \]

5. **Final result:**
The energy of the emitted electrons is:
\[ KE = 1.986 \times 10^{-19} \, \text{J} \]