Work function of lithium and copper are respectively `2.3 eV` and `4.0 eV`. Which one of the metal will be useful for the photoelectric cell working with visible light ?
`(h = 6.6 xx 10^(-34) J - s, c = 3 xx 10^(8) m//s)`

Options

A. lithium
B. Copper
C. Both
D. None of these

Answer

To determine which metal, lithium or copper, is useful for a photoelectric cell working with visible light, we need to compare their work functions and the maximum wavelengths of light that can cause the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding Work Function**:
The work function (φ) is the minimum energy required to remove an electron from the surface of a material. It is given in electron volts (eV).

2. **Given Values**:
- Work function of lithium (φ_Li) = 2.3 eV
- Work function of copper (φ_Cu) = 4.0 eV
- Planck's constant (h) = 6.6 x 10^(-34) J·s
- Speed of light (c) = 3 x 10^(8) m/s
- 1 eV = 1.6 x 10^(-19) J

3. **Calculating Maximum Wavelength (λ₀)**:
The maximum wavelength of light that can cause the photoelectric effect is given by the formula:
\[ λ₀ = \frac{hc}{φ} \]
where φ is the work function in joules.

4. **Convert Work Function to Joules**:
For lithium:
\[ φ_Li = 2.3 \, \text{eV} = 2.3 \times 1.6 \times 10^{-19} \, \text{J} = 3.68 \times 10^{-19} \, \text{J} \]
For copper:
\[ φ_Cu = 4.0 \, \text{eV} = 4.0 \times 1.6 \times 10^{-19} \, \text{J} = 6.4 \times 10^{-19} \, \text{J} \]

5. **Calculate λ₀ for Lithium**:
\[ λ₀(Li) = \frac{(6.6 \times 10^{-34} \, \text{J·s}) \times (3 \times 10^{8} \, \text{m/s})}{3.68 \times 10^{-19} \, \text{J}} \]
\[ λ₀(Li) = \frac{1.98 \times 10^{-25}}{3.68 \times 10^{-19}} \approx 5.38 \times 10^{-7} \, \text{m} = 5380 \, \text{Å} \]

6. **Calculate λ₀ for Copper**:
\[ λ₀(Cu) = \frac{(6.6 \times 10^{-34} \, \text{J·s}) \times (3 \times 10^{8} \, \text{m/s})}{6.4 \times 10^{-19} \, \text{J}} \]
\[ λ₀(Cu) = \frac{1.98 \times 10^{-25}}{6.4 \times 10^{-19}} \approx 3.09 \times 10^{-7} \, \text{m} = 3094 \, \text{Å} \]

7. **Determine Visibility**:
The visible light spectrum ranges approximately from 4000 Å (400 nm) to 7000 Å (700 nm).

- For lithium, λ₀(Li) = 5380 Å, which is within the visible range.
- For copper, λ₀(Cu) = 3094 Å, which is outside the visible range (ultraviolet).

8. **Conclusion**:
Since lithium has a maximum wavelength that falls within the visible range, it is the suitable metal for a photoelectric cell working with visible light.

### Final Answer:
Lithium will be useful for the photoelectric cell working with visible light.

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