When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

Options

A. `1 : 2`
B. `2 : 1`
C. `4 : 1`
D. `1 : 4`

Explanation

To solve the problem, we need to determine the ratio of the work functions of two photoelectric emitters based on the wavelengths of light that can liberate photoelectrons from them.

### Step-by-Step Solution:

1. **Understand the Work Function**:
The work function (φ) is the minimum energy required to liberate an electron from the surface of a material. The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the light.

2. **Calculate the Work Function for Each Emitter**:
- For the first emitter (with wavelength \( \lambda_1 = 300 \, \text{nm} \)):
\[ \phi_1 = \frac{hc}{\lambda_1} \]
- For the second emitter (with wavelength \( \lambda_2 = 600 \, \text{nm} \)):
\[ \phi_2 = \frac{hc}{\lambda_2} \]

3. **Find the Ratio of Work Functions**:
To find the ratio of the work functions \( \frac{\phi_1}{\phi_2} \):
\[ \frac{\phi_1}{\phi_2} = \frac{\frac{hc}{\lambda_1}}{\frac{hc}{\lambda_2}} = \frac{\lambda_2}{\lambda_1} \]
Since \(hc\) cancels out, we are left with the ratio of the wavelengths.

4. **Substitute the Values**:
Substitute the values of the wavelengths:
\[ \frac{\phi_1}{\phi_2} = \frac{600 \, \text{nm}}{300 \, \text{nm}} = 2 \]

5. **Conclusion**:
The ratio of the work functions of the two emitters is:
\[ \frac{\phi_1}{\phi_2} = 2:1 \]

### Final Answer:
The ratio of the work function of the two emitters is \(2:1\).

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