A black body is at temperature of 5780 K. The energy of radiation emitted by the body at wavelength 300 nm is $U_1$, at wavelength 500 nm is $U_2$ and that at 900 nm is $U_3$ respectively. Wien's constant $b = 2.89\times 10^6 \text{nmK}$. This shows that

Question

A star 'A' has radiant power equal to 3 times that of the Sun. The temperature of star 'A' is $ 6000 \, \mathrm{K} $ and that of the Sun is $ 2000 \, \mathrm{K} $. What is the ratio of their radii?

Answer 1

Step 1: Approach
Think of the graph of spectral emission against wavelength.

Step 2: Peak
$\lambda_m=\dfrac{2.89\times10^6}{5780}=500$ nm. The curve climbs from short wavelengths up to 500 nm and then declines.

Step 3: Values
300 nm lies on the rising side, 500 nm is the top, and 900 nm lies on the falling side. So $U_2$ exceeds both $U_1$ and $U_3$.

Step 4: Answer
$U_1<U_2>U_3$, option (C).

Final Answer:
The peak lies at 500 nm, so the energy at 500 nm is greater than at 300 nm and at 900 nm, option (C). \[ \boxed{U_1<U_2>U_3} \]

A black body is at temperature of 5780 K. The energy of radiation emitted by the body at wavelength 300 nm is \(U_1\), at wavelength 500 nm is \(U_2\) and that at 900 nm is \(U_3\) respectively. Wien's constant \(b = 2.89\times 10^6 \text{nmK}\). This shows that

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on radiation

Questions Asked in MHT CET exam