Q135 / 150 · MHT CET 2024 (09 May Shift 2) · Physics

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

  • A
    64 P
  • Correct answer
    B
    256 P
  • C
    4 P
  • D
    16 P

Solution

According to Wien's displacement law, $\lambda_{\text {max }} \mathrm{T}=$ constant $\therefore \quad \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\lambda_{\max _2}}{\lambda_{\max _1}}=\frac{1 / 4 \lambda_0}{\lambda_0}=\frac{1}{4}$ Power radiated for a blackbody, $\mathrm{P}=\sigma \mathrm{AT}^4$ $\begin{aligned} \therefore \quad \frac{\mathrm{P}_1}{\mathrm{P}_2} & =\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\left(\frac{1}{4}\right)^4=\frac{1}{256} \\ \mathrm{P}_2 & =256 \mathrm{P}_1=256 \mathrm{P} \end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions