Question

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

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

Answer 1

Answer: (C)

Given, maximum wavelength,
${\lambda }_m=289.8nm=289.8\times 10^{-9}m$
$=2898\times 10^{-10}m$
Stefan's constant, $\sigma =5.67\times 10^{-8}W{m}^{-2}{K}^{-4}$
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
$E=\sigma Ae{T}^4...(ii)$
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
$I=\sigma {\left(10^4\right)}^4=5.67\times 10^{-8}\times 10^{16}$
$[\because \sigma =5.67\times 10^{-8}\text{(given)] }$
$=5.67\times 10^{8}W{m}^{-2}$