Question 729: Thermodynamics & KTG

Question

The power radiated by a black body is P and it radiates maximum energy around the wavelength $\lambda_0$. Now the temperature of the black body is changed so that it radiates maximum energy around wavelength $\left(\frac{\lambda_0}{2}\right)$. The power radiated by it will now increase by a factor of

Accepted Answer

According to Wien's Law, the product of the wavelength at which a black body radiates maximum energy ($\lambda_m$) and its temperature (T) is a constant:
$ \lambda_m T = 	ext{constant} $
If initially the black body radiates maximum energy at wavelength $\lambda_0$ and temperature $T_1$, and then it changes to radiate maximum energy at $\frac{\lambda_0}{2}$, the new temperature $T_2$ can be found as:
$ \lambda_{m_1} T_1 = \lambda_{m_2} T_2 $
Substituting the given values:
$ T_2 = \frac{\lambda_{m_1}}{\lambda_{m_2}} T_1 = \frac{\lambda_0}{\frac{\lambda_0}{2}} T_1 = 2 T_1 \quad 	ext{... (i)} $
According to the Stefan-Boltzmann Law, the power radiated by a black body is proportional to the fourth power of its temperature:
$ P \propto T^4 $
For the new power $P_2$ at temperature $T_2$, we have:
$ \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}ight)^4 $
Substituting for $T_2$ from equation (i):
$ \frac{P_2}{P_1} = \left(\frac{2 T_1}{T_1}ight)^4 = 16 $
Thus, the power radiated by the black body increases by a factor of 16.

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2

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8

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16

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32

Thermodynamics & KTG

The power radiated by a black body is P and it radiates maximum energy around the wavelength \lambda_0. Now the temperature of the black body is changed so that it radiates maximum energy around wavelength \left(\frac{\lambda_0}{2}\right). The power radiated by it will now increase by a factor of

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