MHT CET · Physics · Thermal Properties of Matter
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
- A 2
- B 8
- C 16
- D 32
Answer & Solution
Correct Answer
(C) 16
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& \text { According to Wien’s law, } \\
& \lambda_{\mathrm{m}} \mathrm{~T}=\text { constant }. \\
& \lambda_{m_1} T_1=\lambda_{m_2} T_2 \\
\therefore \quad & T_2=\frac{\lambda_{m_1}}{\lambda_{m_2}} T_1=\frac{\lambda_0}{\left(\frac{\lambda_0}{2}\right)} \times T_1=2 T_1...(i)
\end{array}\)
From Stefan-Boltzmann law, \(\mathrm{P} \propto \mathrm{T}^4\)
\(\therefore \quad \frac{P_2}{P_r}=\left(\frac{T_2}{T_1}\right)^4\).
\(\therefore \quad \frac{P_2}{P_1}=\left(\frac{2 T_1}{T_1}\right)^4=16\)
& \text { According to Wien’s law, } \\
& \lambda_{\mathrm{m}} \mathrm{~T}=\text { constant }. \\
& \lambda_{m_1} T_1=\lambda_{m_2} T_2 \\
\therefore \quad & T_2=\frac{\lambda_{m_1}}{\lambda_{m_2}} T_1=\frac{\lambda_0}{\left(\frac{\lambda_0}{2}\right)} \times T_1=2 T_1...(i)
\end{array}\)
From Stefan-Boltzmann law, \(\mathrm{P} \propto \mathrm{T}^4\)
\(\therefore \quad \frac{P_2}{P_r}=\left(\frac{T_2}{T_1}\right)^4\).
\(\therefore \quad \frac{P_2}{P_1}=\left(\frac{2 T_1}{T_1}\right)^4=16\)
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