Heat and Thermodynamics 3 Question 18

18. A black body is at a temperature of $2880 K$. The energy of radiation emitted by this body with wavelength between $499 nm$ and $500 nm$ is $U _1$, between $999 nm$ and $1000 nm$ is $U _2$ and between $1499 nm$ and $1500 nm$ is $U _3$. The Wien constant, $b=2.88 \times 10^{6} nm-K$. Then,

(1998, 2M)

(a) $U _1=0$

(b) $U _3=0$

(c) $U _1>U _2$

(d) $U _2>U _1$

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Answer:

Correct Answer: 18. (d)

Solution:

  1. Wien’s displacement law is

$$ \begin{aligned} & \lambda _m T & =b \quad(b=\text { Wien’s constant }) \\ \therefore & \lambda _m & =\frac{b}{T}=\frac{2.88 \times 10^{6} nm-K}{2880 K} \\ \therefore & \lambda & =1000 nm \end{aligned} $$

Energy distribution with wavelength will be as follows :

From the graph it is clear that

$$ U _2>U _1 \quad \text { (In fact } U _2 \text { is maximum) } $$



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