The work function of a metal is in the range of 2 eV to 5 eV. Find which of the following wavelength of light cannot be used for photoelectric effect. (Consider, Planck's constant `=4xx10^(-15)" eV - s, velocity of light "=3xx10^(8)ms^(-1))`

Options

A. 510 nm
B. 650 nm
C. 400 nm
D. 570 nm

Answer

To determine which wavelength of light cannot be used for the photoelectric effect, we need to calculate the energy of photons corresponding to each given wavelength and compare it with the work function range of the metal (2 eV to 5 eV). The photoelectric effect will occur if the energy of the photons is greater than or equal to the work function.

### Step-by-Step Solution:

1. **Understand the Relationship Between Energy and Wavelength**:
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 4 \times 10^{-15} \, \text{eV} \cdot \text{s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

Alternatively, we can use the simplified formula:
\[ E = \frac{1240}{\lambda} \]
where \( \lambda \) is in nanometers and \( E \) is in electron volts.

2. **Calculate the Energy for Each Wavelength**:
Let's calculate the energy for each of the given wavelengths.

- **For \( \lambda_1 = 510 \, \text{nm} \)**:
\[ E_1 = \frac{1240}{510} \approx 2.43 \, \text{eV} \]

- **For \( \lambda_2 = 650 \, \text{nm} \)**:
\[ E_2 = \frac{1240}{650} \approx 1.90 \, \text{eV} \]

- **For \( \lambda_3 = 400 \, \text{nm} \)**:
\[ E_3 = \frac{1240}{400} \approx 3.10 \, \text{eV} \]

- **For \( \lambda_4 = 570 \, \text{nm} \)**:
\[ E_4 = \frac{1240}{570} \approx 2.17 \, \text{eV} \]

3. **Compare the Energies with the Work Function**:
The work function of the metal is in the range of 2 eV to 5 eV. We need to check which energies are less than 2 eV:
- \( E_1 = 2.43 \, \text{eV} \) (can cause photoelectric effect)
- \( E_2 = 1.90 \, \text{eV} \) (cannot cause photoelectric effect)
- \( E_3 = 3.10 \, \text{eV} \) (can cause photoelectric effect)
- \( E_4 = 2.17 \, \text{eV} \) (can cause photoelectric effect)

4. **Conclusion**:
The wavelength of light that cannot be used for the photoelectric effect is \( \lambda_2 = 650 \, \text{nm} \) since its energy \( E_2 = 1.90 \, \text{eV} \) is less than the minimum work function of 2 eV.

### Final Answer:
The wavelength of light that cannot be used for the photoelectric effect is **650 nm**.