A black body at a temperature of 1640,K has t | vedclass

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

Question

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

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(C) According to Wien's displacement law,the product of the temperature $T$ and the wavelength corresponding to maximum emission $\lambda_m$ is a constant:$\lambda_m T = b$ (constant)Therefore,$\lambda_{m1} T_1 = \lambda_{m2} T_2$.Given:$T_1 = 1640\,K$$\lambda_{m1} = 1.75\,\mu m$$\lambda_{m2} = 14.35\,\mu m$We need to find $T_2$:$T_2 = T_1 	imes \frac{\lambda_{m1}}{\lambda_{m2}}$$T_2 = 1640 	imes \frac{1.75}{14.35}$$T_2 \approx 1640 	imes 0.12195$$T_2 \approx 200\,K$.

$A$ black body at a temperature of $1640\,K$ has the wavelength corresponding to maximum emission equal to $1.75\,\mu m$. Assuming the moon to be a perfectly black body,the temperature of the moon,if the wavelength corresponding to maximum emission is $14.35\,\mu m$,is.......$K$.

Similar Questions

If $x = \tan 15^{\circ}$,$y = \operatorname{cosec} 75^{\circ}$ and $z = 4 \sin 18^{\circ}$,then :

If $A$ and $B$ are two independent events such that $P(B)=\frac{2}{7}$ and $P(A \cup B^c)=0.8$, then $P(A)$ is equal to:

An image formed by a concave mirror of focal length $6 \ cm$ is $3$ times the size of the object. What is the distance of the object from the mirror in $cm$?

You are given a fairly old piece of dicot stem and a dicot root. Which of the following anatomical structures will you use to distinguish between the two?

From which plant is cocaine obtained?

Vedclass Test Series

Exam Paper Generator

Online Exam Module