When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitted , 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 find 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. **Understanding Work Function**:
The work function (Φ) of a material is the minimum energy required to liberate an electron from the surface of that material. It can be expressed in terms of the threshold wavelength (λ₀) as:
\[ \Phi = \frac{hc}{\lambda_0} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Identify the Wavelengths**:
We have two wavelengths:
- For emitter 1 (λ₁): 300 nm
- For emitter 2 (λ₂): 600 nm

3. **Calculate Work Functions**:
Using the formula for work function:
- For emitter 1:
\[ \Phi_1 = \frac{hc}{\lambda_1} = \frac{hc}{300 \text{ nm}} \]
- For emitter 2:
\[ \Phi_2 = \frac{hc}{\lambda_2} = \frac{hc}{600 \text{ nm}} \]

4. **Finding the Ratio of Work Functions**:
To find the ratio of the work functions, we can divide Φ₁ by Φ₂:
\[ \frac{\Phi_1}{\Phi_2} = \frac{\frac{hc}{\lambda_1}}{\frac{hc}{\lambda_2}} = \frac{\lambda_2}{\lambda_1} \]

5. **Substituting the Values**:
Substitute the values of λ₁ and λ₂:
\[ \frac{\Phi_1}{\Phi_2} = \frac{600 \text{ nm}}{300 \text{ nm}} = 2 \]

6. **Final Ratio**:
Therefore, the ratio of the work functions of the two emitters is:
\[ \Phi_1 : \Phi_2 = 2 : 1 \]

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

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