When sunlight is scattered by atmospheric atoms and molecules the amount of scattering of light of wavelength 880 nm is A. Then the , the amount of scattering of light of wavelength 330 nm is approximately

Options

A. `10A`
B. `20A`
C. `40A`
D. `50.5A`

Answer

To solve the problem of how much light of wavelength 330 nm is scattered compared to light of wavelength 880 nm, we can use Rayleigh's scattering law. The law states that the intensity of scattered light is inversely proportional to the fourth power of the wavelength.

### Step-by-Step Solution:

1. **Understanding Rayleigh's Scattering Law**:
The intensity of scattered light (I) is related to the wavelength (λ) by the formula:
\[ I \propto \frac{1}{\lambda^4} \]
This means that if we have two different wavelengths, we can express their intensities as:
\[ \frac{I_1}{I_2} = \frac{\lambda_2^4}{\lambda_1^4} \]

2. **Identifying the Variables**:
- Let \( I_1 \) be the intensity of light with wavelength 880 nm, which is given as \( A \).
- Let \( I_2 \) be the intensity of light with wavelength 330 nm (the value we want to find).
- The wavelengths are:
- \( \lambda_1 = 880 \, \text{nm} \)
- \( \lambda_2 = 330 \, \text{nm} \)

3. **Setting Up the Equation**:
Using the relationship from Rayleigh's law, we can set up the equation:
\[ \frac{A}{I_2} = \frac{(330)^4}{(880)^4} \]

4. **Calculating the Ratio**:
This can be simplified as:
\[ I_2 = A \cdot \frac{(880)^4}{(330)^4} \]
We can express this as:
\[ I_2 = A \cdot \left(\frac{880}{330}\right)^4 \]

5. **Calculating the Numerical Values**:
First, calculate the ratio:
\[ \frac{880}{330} = \frac{88}{33} = \frac{8}{3} \]
Now raise it to the power of 4:
\[ \left(\frac{8}{3}\right)^4 = \frac{8^4}{3^4} = \frac{4096}{81} \]

6. **Final Calculation**:
Now substituting back into the equation for \( I_2 \):
\[ I_2 = A \cdot \frac{4096}{81} \]
This gives us the final intensity of light with wavelength 330 nm in terms of \( A \).

### Conclusion:
Thus, the amount of scattering of light of wavelength 330 nm is approximately:
\[ I_2 \approx 50.56 A \]