Two spherical bodies A (radius 6cm) and B (radius 18cm) are at temperature `T_(1)` and `T_(2)` respectively The maximum intensity in the emission spectrum of A is at `500 nm` and in that of `B` is at `1500nm` considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of `B` . ?

Explanation

To find the ratio of the rate of total energy radiated by two spherical bodies A and B, we will use Wien's law and Stefan-Boltzmann law.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of body A, \( R_A = 6 \, \text{cm} \)
- Radius of body B, \( R_B = 18 \, \text{cm} \)
- Wavelength of maximum intensity for A, \( \lambda_A = 500 \, \text{nm} \)
- Wavelength of maximum intensity for B, \( \lambda_B = 1500 \, \text{nm} \)

2. **Use Wien's Displacement Law:**
- According to Wien's law, the product of the temperature \( T \) and the wavelength \( \lambda \) at which the emission is maximum is constant:
\[ \lambda_A T_A = \lambda_B T_B \]
- Rearranging gives us:
\[ \frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A} \]
- Substituting the values:
\[ \frac{T_A}{T_B} = \frac{1500 \, \text{nm}}{500 \, \text{nm}} = 3 \]

3. **Apply Stefan-Boltzmann Law:**
- The power (rate of energy radiated) for a black body is given by:
\[ P = \sigma A T^4 \]
- Where \( \sigma \) is the Stefan-Boltzmann constant and \( A \) is the surface area.
- For a sphere, the surface area \( A \) is given by \( A = 4\pi R^2 \).

4. **Calculate the Ratio of Power:**
- The power radiated by A and B can be expressed as:
\[ P_A = \sigma (4\pi R_A^2) T_A^4 \]
\[ P_B = \sigma (4\pi R_B^2) T_B^4 \]
- The ratio \( \frac{P_A}{P_B} \) becomes:
\[ \frac{P_A}{P_B} = \frac{R_A^2 T_A^4}{R_B^2 T_B^4} \]

5. **Substituting Values:**
- Substitute \( R_A = 6 \, \text{cm} \) and \( R_B = 18 \, \text{cm} \):
\[ \frac{P_A}{P_B} = \frac{(6)^2 T_A^4}{(18)^2 T_B^4} \]
- This simplifies to:
\[ \frac{P_A}{P_B} = \frac{36 T_A^4}{324 T_B^4} = \frac{1}{9} \cdot \frac{T_A^4}{T_B^4} \]

6. **Using the Temperature Ratio:**
- From step 2, we found \( \frac{T_A}{T_B} = 3 \):
\[ \frac{T_A^4}{T_B^4} = 3^4 = 81 \]
- Therefore, substituting this back gives:
\[ \frac{P_A}{P_B} = \frac{1}{9} \cdot 81 = 9 \]

### Final Answer:
The ratio of the rate of total energy radiated by A to that of B is:
\[ \frac{P_A}{P_B} = 9 \]