A particular star (assuming it as a black bod | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a const

Vedclass · Accepted Answer

$58$

Vedclass · Answer

(B) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant $(b)$.$\lambda_m T = b$Given:$T = 5 	imes 10^4 \ K$$b = 0.0029 \ mK$Substituting the values:$\lambda_m = \frac{b}{T} = \frac{0.0029}{5 	imes 10^4}$$\lambda_m = \frac{2.9 	imes 10^{-3}}{5 	imes 10^4} = 0.58 	imes 10^{-7} \ m$$\lambda_m = 58 	imes 10^{-9} \ m$Since $1 \ nm = 10^{-9} \ m$,we get:$\lambda_m = 58 \ nm$.

$A$ particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^4 \ K$. The wavelength in nanometers at which its radiation becomes maximum is $(b = 0.0029 \ mK)$.

Similar Questions

The following graph represents the radiant power versus wavelength of a black body. The area under the curve represents:

Write the temperature of the surfaces of the Moon and the Sun according to Wien's displacement law.

The colour of a shining bright star is an indication of its

$A$ black body at $200 K$ is found to emit maximum energy at a wavelength of $14 \mu m$. When its temperature is raised to $1000 K$,the wavelength at which maximum energy is emitted is:

The wavelength of radiation emitted by a body depends upon

Vedclass Test Series

Exam Paper Generator

Online Exam Module