The maximum kinetic energy of electron if wavelenght of incident electromagnetic wave is `260 nm` and cut-off wvelenght is `380 nm` given `hc = 1237nm.eV` is

Options

A. `1.5eV`
B. `6.4eV`
C. `10eV`
D. `None of these

Answer

To find the maximum kinetic energy of the electron when the wavelength of the incident electromagnetic wave is 260 nm and the cut-off wavelength is 380 nm, we can use the photoelectric effect formula. The maximum kinetic energy (K.E.) of the emitted electron can be calculated using the equation:

\[ K.E. = h\nu - \phi \]

Where:
- \( h\nu \) is the energy of the incident photon,
- \( \phi \) is the work function of the material, which can be expressed in terms of the cut-off wavelength \( \lambda_0 \).

### Step-by-Step Solution:

1. **Calculate the energy of the incident photon (\( h\nu \))**:
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Given \( \lambda = 260 \, \text{nm} \) and \( hc = 1237 \, \text{nm.eV} \):
\[ E = \frac{1237 \, \text{nm.eV}}{260 \, \text{nm}} = \frac{1237}{260} \approx 4.75 \, \text{eV} \]

2. **Calculate the work function (\( \phi \))**:
The work function can be calculated using the cut-off wavelength:
\[ \phi = \frac{hc}{\lambda_0} \]
Where \( \lambda_0 = 380 \, \text{nm} \):
\[ \phi = \frac{1237 \, \text{nm.eV}}{380 \, \text{nm}} = \frac{1237}{380} \approx 3.25 \, \text{eV} \]

3. **Calculate the maximum kinetic energy (K.E.)**:
Now, substituting the values of \( h\nu \) and \( \phi \) into the kinetic energy formula:
\[ K.E. = h\nu - \phi = 4.75 \, \text{eV} - 3.25 \, \text{eV} = 1.50 \, \text{eV} \]

### Final Answer:
The maximum kinetic energy of the electron is:
\[ \boxed{1.5 \, \text{eV}} \]