Question 769: Thermodynamics & KTG

Question

A black body radiates power ' P ' and maximum energy is radiated by it at a wavelength $\lambda_0$. The temperature of the black body is now so changed that it radiates maximum energy at the wavelength $\frac{\lambda_0}{4}$. The power radiated by it at new temperature is

Accepted Answer

According to Wien's Displacement Law:
$ \lambda_{	ext{max}} \cdot T = 	ext{constant} $
Given this, we can relate the initial and final temperatures using the wavelengths:
$ \frac{T_1}{T_2} = \frac{\lambda_{	ext{max, 2}}}{\lambda_{	ext{max, 1}}} = \frac{\frac{\lambda_0}{4}}{\lambda_0} = \frac{1}{4} $
The power radiated by a black body is given by Stefan-Boltzmann law:
$ P = \sigma A T^4 $
We need to determine the ratio of the powers:
$ \frac{P_1}{P_2} = \left(\frac{T_1}{T_2}ight)^4 = \left(\frac{1}{4}ight)^4 = \frac{1}{256} $
Solving for $ P_2 $ gives:
$ P_2 = 256 \cdot P_1 = 256 \cdot P $
Thus, the power radiated at the new temperature is $ 256P $.

Answer

64 P

Answer

256 P

Answer

4 P

Answer

16 P

Thermodynamics & KTG

A black body radiates power ' P ' and maximum energy is radiated by it at a wavelength \lambda_0. The temperature of the black body is now so changed that it radiates maximum energy at the wavelength \frac{\lambda_0}{4}. The power radiated by it at new temperature is

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