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

The strongest Bronsted base among the following anions is:

If $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$,$l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \pi / 2} \frac{\cos x}{x - \pi / 2}$,then:

Where is the International Rice Research Institute $(IRRI)$ located?

$A$ prism of refractive index $\mu$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is $A$,then the value of $A$ in terms of $\mu$ is :

Two seconds after projection,a projectile is travelling in a direction inclined at $30^o$ to the horizontal. After one more second,it is travelling horizontally. What is the magnitude and direction of its velocity at the initial point?

Vedclass Test Series

Exam Paper Generator

Online Exam Module