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, we will use Rayleigh's Law of Scattering, which states that the intensity of scattered light is inversely proportional to the fourth power of the wavelength of the light.

### Step-by-Step Solution:

1. **Understand Rayleigh's Law of Scattering**:
According to Rayleigh's law, the intensity of scattering (I) is inversely proportional to the fourth power of the wavelength (λ):
\[ I \propto \frac{1}{\lambda^4} \]

2. **Set Up the Ratio**:
We can express the relationship between the intensities of scattering at two different wavelengths (λ1 and λ2) as:
\[ \frac{I_1}{I_2} = \left(\frac{\lambda_2}{\lambda_1}\right)^4 \]
where \(I_1\) is the intensity of scattering at wavelength \(λ_1\) (880 nm), and \(I_2\) is the intensity of scattering at wavelength \(λ_2\) (330 nm).

3. **Substitute Known Values**:
Let \(I_1 = A\), \(λ_1 = 880 \, nm\), and \(λ_2 = 330 \, nm\):
\[ \frac{A}{I_2} = \left(\frac{330}{880}\right)^4 \]

4. **Calculate the Ratio**:
First, simplify the fraction:
\[ \frac{330}{880} = \frac{33}{88} = \frac{3}{8} \]
Now, substitute this back into the equation:
\[ \frac{A}{I_2} = \left(\frac{3}{8}\right)^4 \]

5. **Calculate \((\frac{3}{8})^4\)**:
\[ \left(\frac{3}{8}\right)^4 = \frac{3^4}{8^4} = \frac{81}{4096} \]

6. **Rearranging for \(I_2\)**:
Now, rearranging the equation gives:
\[ I_2 = A \cdot \frac{4096}{81} \]

7. **Final Calculation**:
Now, calculate the value of \(I_2\):
\[ I_2 = A \cdot \frac{4096}{81} \approx 50.557A \]

### Conclusion:
The amount of scattering of light of wavelength 330 nm is approximately \(50.557A\).