Light of wavelength 4000 `Å` falls on a photosensitive plate with photoelectric work function 1.96 eV. The K.E. of photoelectrons emitted will be :

Options

A. 1.148 eV
B. 1.152 eV
C. 1.156 eV
D. 1.158 eV

Answer

To solve the problem, we need to calculate the kinetic energy (K.E.) of the photoelectrons emitted when light of wavelength 4000 Å falls on a photosensitive plate with a work function of 1.96 eV. We will use the photoelectric effect equation to find the answer.

### Step-by-Step Solution:

1. **Convert Wavelength to Meters**:
The wavelength given is 4000 Å (angstroms). We need to convert this to meters for our calculations.
\[ 1 \text{ Å} = 10^{-10} \text{ m} \]
Therefore,
\[ \text{Wavelength} = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} = 4 \times 10^{-7} \text{ m} \]

2. **Calculate the Energy of the Incident Photon**:
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \text{ J s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \text{ m/s} \)
- \( \lambda \) = \( 4 \times 10^{-7} \text{ m} \)

Plugging in the values:
\[ E = \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{4 \times 10^{-7} \text{ m}} \]
\[ E = \frac{1.9878 \times 10^{-25} \text{ J m}}{4 \times 10^{-7} \text{ m}} = 4.9695 \times 10^{-19} \text{ J} \]

3. **Convert Energy from Joules to Electron Volts**:
To convert the energy from Joules to electron volts (eV), we use the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Thus,
\[ E = \frac{4.9695 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 3.10 \text{ eV} \]

4. **Use the Photoelectric Equation**:
According to Einstein's photoelectric equation, the maximum kinetic energy (K.E.) of the emitted photoelectrons is given by:
\[ K.E. = E - \phi \]
where \( \phi \) is the work function of the material (1.96 eV in this case).

Substituting the values:
\[ K.E. = 3.10 \text{ eV} - 1.96 \text{ eV} = 1.14 \text{ eV} \]

5. **Final Answer**:
The kinetic energy of the photoelectrons emitted will be approximately:
\[ K.E. \approx 1.14 \text{ eV} \]