When light of wavelength `300 nm` (nanometer) falls on a photoelectric emitter, photoelectrons are emitted. In another emitter however light of `600 nm` wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters ?

Options

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

Answer

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 cause photoemission.

### Step-by-Step Solution:

1. **Understand Work Function**: The work function (φ) of a material is the minimum energy required to remove an electron from the surface of that material. It can be expressed in terms of the threshold frequency (ν₀) as:
\[ \phi = h \cdot \nu_0 \]
where \(h\) is Planck's constant.

2. **Relate Frequency to Wavelength**: The frequency (ν) of light is related to its wavelength (λ) by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \(c\) is the speed of light.

3. **Express Work Function in Terms of Wavelength**: We can substitute the frequency in the work function equation:
\[ \phi = h \cdot \frac{c}{\lambda_0} \]
where \(\lambda_0\) is the wavelength corresponding to the threshold frequency.

4. **Calculate Work Functions for Both Emitters**:
- For the first emitter (λ₁ = 300 nm):
\[ \phi_1 = h \cdot \frac{c}{\lambda_1} = h \cdot \frac{c}{300 \, \text{nm}} \]
- For the second emitter (λ₂ = 600 nm):
\[ \phi_2 = h \cdot \frac{c}{\lambda_2} = h \cdot \frac{c}{600 \, \text{nm}} \]

5. **Find the Ratio of Work Functions**:
\[ \frac{\phi_1}{\phi_2} = \frac{h \cdot \frac{c}{\lambda_1}}{h \cdot \frac{c}{\lambda_2}} = \frac{\lambda_2}{\lambda_1} \]
Since \(h\) and \(c\) cancel out, we have:
\[ \frac{\phi_1}{\phi_2} = \frac{600 \, \text{nm}}{300 \, \text{nm}} = 2 \]

6. **Final Ratio**: Therefore, the ratio of the work functions is:
\[ \phi_1 : \phi_2 = 2 : 1 \]

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

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