Two spherical black bodies A and B are emitting radiations at same rate. The radius of B is doubled keeping radius of A fixed. The wavelength corresponding to maximum intensity becomes half for A white it ramains same for B. Then, the ratio of rate radiation energy emitted by a and is B is

Options

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

Explanation

To solve the problem, we need to analyze the information given about two spherical black bodies A and B, their rates of radiation, and the changes in their physical properties.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Both bodies A and B are emitting radiation at the same rate initially.
- The radius of body B is doubled while the radius of body A remains fixed.
- The wavelength corresponding to maximum intensity for body A becomes half, while it remains the same for body B.

2. **Using the Stefan-Boltzmann Law:**
The rate of radiation energy emitted by a black body is given by:
\[ E = \sigma A T^4 \]
where \(E\) is the rate of radiation energy, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the temperature.

3. **Surface Area Calculation:**
The surface area \(A\) of a sphere is given by:
\[ A = 4\pi r^2 \]
For body A:
\[ A_A = 4\pi R_A^2 \]
For body B (with doubled radius):
\[ A_B = 4\pi (2R_B)^2 = 16\pi R_B^2 \]

4. **Temperature Changes:**
Since the wavelength corresponding to maximum intensity is inversely proportional to temperature (\(\lambda \propto \frac{1}{T}\)):
- For body A, if the wavelength becomes half, the temperature doubles:
\[ T_A' = 2T_A \]
- For body B, the temperature remains the same:
\[ T_B' = T_B \]

5. **Calculating the Rate of Radiation for Each Body:**
For body A:
\[ E_A = \sigma (4\pi R_A^2) (2T_A)^4 = \sigma (4\pi R_A^2) \cdot 16T_A^4 = 64\sigma \pi R_A^2 T_A^4 \]
For body B:
\[ E_B = \sigma (16\pi R_B^2) (T_B)^4 = 16\sigma \pi R_B^2 T_B^4 \]

6. **Finding the Ratio of Rates of Radiation:**
Now we can find the ratio of the rates of radiation:
\[ \frac{E_A}{E_B} = \frac{64\sigma \pi R_A^2 T_A^4}{16\sigma \pi R_B^2 T_B^4} \]
Simplifying this gives:
\[ \frac{E_A}{E_B} = \frac{64 R_A^2 T_A^4}{16 R_B^2 T_B^4} = 4 \cdot \frac{R_A^2 T_A^4}{R_B^2 T_B^4} \]

7. **Using the Initial Condition:**
From the initial condition where both bodies emitted radiation at the same rate:
\[ R_A^2 T_A^4 = R_B^2 T_B^4 \]
Thus, we can substitute this into our ratio:
\[ \frac{E_A}{E_B} = 4 \cdot 1 = 4 \]

### Final Answer:
The ratio of the rate of radiation energy emitted by A to that emitted by B is:
\[ \frac{E_A}{E_B} = 4 \]