Question

The maximum wavelength of radiation emitted by a star is $289.8 \,nm$. Then intensity of radiation for the star is (Given : Stefan's constant $= 5.67 \times 10^{-8}\, Wm^{-2}K^{-4}$ , Wien’s constant, $b = 2898\, \mu\, mK$)

• A. $5.67 \times 10^{-12} Wm^{-2}$
• B. $10.67 \times 10^{14} Wm^{-2}$
• C. $5.67 \times 10^{8} Wm^{-2}$
• D. $10.67 \times 10^{7} Wm^{-2}$

Answer 1

Answer: (C)

Given, maximum wavelength,
$\lambda_{m} =289.8\, nm =289.8 \times 10^{-9} m$
$=2898 \times 10^{-10} m$
Stefan's constant, $\sigma=5.67 \times 10^{-8} Wm ^{-2} K ^{-4} $
Wien's constant, $b=2898 \mu mK =2898 \times 10^{-6} mK$
According to Wien's displacement law, the maximum wavelength is given by
$\lambda_{m}=\frac{b}{T} \Rightarrow T=\frac{b}{\lambda_{m}}\,\,\,\,\,\,\,...(i)$
Substituting given values in Eq. (i), we get
$T=\frac{2898 \times 10^{-6}}{2898 \times 10^{-10}}=10^{4} K \,\,\,\,\,\,\, ...(ii)$
According to Stefan's law, the energy radiated from a source is given by
$E=\sigma A e T^{4} \,\,\,\,\,\,\,\,\,...(ii)$
where, $A=$ area of source
$e=$ emissivity (value between 0 to 1$)$
The intensity of radiations emitted is equal to energy radiated from a given surface area, i.e.,
$I=\frac{E}{A}=\sigma e T^{4}$ [from Eq. (iii)]
As $e$ is very small, so
$I=\sigma T^{4} \,\,\,\,\,\,\,...(iv)$
Substituting the value of $T$ from Eq.(ii) in Eq. (iv), we get
$I=\sigma\left(10^{4}\right)^{4} =5.67 \times 10^{-8} \times 10^{16}$
$\left[\because \sigma=5.67 \times 10^{-8}\right. \text {(given)] }$
$=5.67 \times 10^{8} Wm ^{-2}$