The black body spectrum of an object `O_(1)` is such that its radiant intensity (i.e. intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object `O_(2)` has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source `O_(1)` to that of source `O_(2)` is

Options

A. `1:81`
B. `1:9`
C. `9:1`
D. `81:1`

Explanation

To solve the problem, we will use Wien's Displacement Law and Stefan-Boltzmann Law to find the ratio of power emitted per unit area by the two objects, O1 and O2.

### Step-by-Step Solution:

1. **Identify the Wavelengths:**
- For object O1, the maximum radiant intensity is at \( \lambda_{m1} = 200 \, \text{nm} \).
- For object O2, the maximum radiant intensity is at \( \lambda_{m2} = 600 \, \text{nm} \).

2. **Use Wien's Displacement Law:**
- Wien's Displacement Law states that the wavelength at which the intensity is maximum is inversely proportional to the temperature of the black body:
\[ \lambda_{m} T = b \]
where \( b \) is Wien's displacement constant (\( b \approx 2.898 \times 10^{-3} \, \text{m K} \)).
- For O1:
\[ T_1 = \frac{b}{\lambda_{m1}} = \frac{b}{200 \times 10^{-9} \, \text{m}} \]
- For O2:
\[ T_2 = \frac{b}{\lambda_{m2}} = \frac{b}{600 \times 10^{-9} \, \text{m}} \]

3. **Calculate the Temperatures:**
- We can express the ratio of the temperatures:
\[ \frac{T_1}{T_2} = \frac{\lambda_{m2}}{\lambda_{m1}} = \frac{600 \, \text{nm}}{200 \, \text{nm}} = 3 \]

4. **Apply Stefan-Boltzmann Law:**
- The power emitted per unit area (E) by a black body is given by:
\[ E = \sigma T^4 \]
- Therefore, the ratio of power emitted per unit area by O1 to that of O2 is:
\[ \frac{E_1}{E_2} = \left(\frac{T_1}{T_2}\right)^4 = 3^4 = 81 \]

5. **Final Ratio:**
- Thus, the ratio of power emitted per unit area by source O1 to that of source O2 is:
\[ \frac{E_1}{E_2} = 81 : 1 \]

### Conclusion:
The ratio of power emitted per unit area by source O1 to that of source O2 is \( 81 : 1 \).