Work function of a metal is 3.0 eV. It is illuminated by a light of wavelength `3xx10^(-7)` m. Calculate the maximum energy of the electron.

Options

A. 4.14 eV
B. 2.14 eV
C. 3.16 eV
D. 1.16 eV

Answer

To solve the problem, we need to calculate the maximum energy of the electron when a metal with a work function of 3.0 eV is illuminated by light of wavelength \(3 \times 10^{-7}\) m. We will use the photoelectric effect equation to find the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Work function (\(\phi\)) = 3.0 eV
- Wavelength (\(\lambda\)) = \(3 \times 10^{-7}\) 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) = \(4.1357 \times 10^{-15}\) eV·s
- \(c\) (speed of light) = \(3 \times 10^8\) m/s

First, we need to convert \(h\) and \(c\) into compatible units. We can use:
\[ hc = 1240 \text{ eV·nm} \]
Since \(1 \text{ nm} = 10^{-9} \text{ m}\), we can convert the wavelength into nanometers:
\[ \lambda = 3 \times 10^{-7} \text{ m} = 3 \times 10^{2} \text{ nm} = 300 \text{ nm} \]

Now, substituting the values into the energy equation:
\[ E = \frac{1240 \text{ eV·nm}}{300 \text{ nm}} = \frac{1240}{300} \text{ eV} \approx 4.1333 \text{ eV} \]

3. **Calculate the Maximum Kinetic Energy of the Electron:**
The maximum kinetic energy (K.E.) of the emitted electron can be calculated using the photoelectric effect equation:
\[ K.E. = E - \phi \]
Substituting the values we found:
\[ K.E. = 4.1333 \text{ eV} - 3.0 \text{ eV} = 1.1333 \text{ eV} \]

4. **Final Result:**
The maximum energy of the electron is approximately:
\[ K.E. \approx 1.13 \text{ eV} \]

### Conclusion:
The maximum energy of the electron emitted from the metal when illuminated by light of wavelength \(3 \times 10^{-7}\) m is approximately \(1.13 \text{ eV}\).