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

An oscillator is nothing but an amplifier with

The correct order of second $I.E.$ of $C, N, O$ and $F$ is:

Which cell organelle can be identified by Sudan Black?

Evolutionary classification is called ........

The base current in a transistor circuit changes from $45 \mu A$ to $140 \mu A$. Accordingly,the collector current changes from $0.2 \text{ mA}$ to $4.0 \text{ mA}$. The current gain is

Vedclass Test Series

Exam Paper Generator

Online Exam Module