When sunlight is scattered by atmospheric atoms and molecules, the amount of scattering of light of wavelength 440 nm is A. The amount of scattering for the light of wavelength 660 nm is approximately

Options

A. `4/9`A
B. 2.25 A
C. 1.5 A
D. `A/6`

Explanation

To solve the problem of the scattering of light of different wavelengths, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Law of Scattering**:
The intensity of scattered light is inversely proportional to the fourth power of the wavelength. This can be expressed mathematically as:
\[ I \propto \frac{1}{\lambda^4} \]
where \( I \) is the intensity of scattered light and \( \lambda \) is the wavelength.

2. **Set Up the Ratio**:
We can set up a ratio of the intensities for two different wavelengths:
\[ \frac{I_1}{I_2} = \left(\frac{\lambda_2}{\lambda_1}\right)^4 \]
Here, \( I_1 \) is the intensity for the wavelength \( \lambda_1 = 440 \, \text{nm} \) and \( I_2 \) is the intensity for the wavelength \( \lambda_2 = 660 \, \text{nm} \).

3. **Substitute the Known Values**:
We know that \( I_1 = A \) (the intensity for 440 nm) and we need to find \( I_2 \):
\[ \frac{A}{I_2} = \left(\frac{660}{440}\right)^4 \]

4. **Simplify the Wavelength Ratio**:
Calculate the ratio of the wavelengths:
\[ \frac{660}{440} = \frac{66}{44} = \frac{3}{2} \]

5. **Raise the Ratio to the Fourth Power**:
Now raise the ratio to the fourth power:
\[ \left(\frac{3}{2}\right)^4 = \frac{3^4}{2^4} = \frac{81}{16} \]

6. **Set Up the Equation**:
Substitute this back into the intensity ratio:
\[ \frac{A}{I_2} = \frac{81}{16} \]

7. **Cross Multiply to Solve for \( I_2 \)**:
Rearranging gives:
\[ I_2 = \frac{16A}{81} \]

8. **Approximate the Result**:
To find an approximate value, we can simplify:
\[ I_2 \approx \frac{A}{6} \]
(since \( \frac{16}{81} \) is approximately \( \frac{1}{6} \)).

### Final Answer:
Thus, the amount of scattering for the light of wavelength 660 nm is approximately:
\[ I_2 \approx \frac{A}{6} \]