Question

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

• A. \(U_1 < U_2 < U_3\)
• B. \(U_1 > U_2 > U_3\)
• C. \(U_1 < U_2 > U_3\)
• D. \(U_1 > U_2 < U_3\)

Hint:

Find the wavelength of peak emission with Wien law and compare each given wavelength with it.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A black body radiates the most energy at the wavelength \(\lambda_m=\dfrac bT\). The spectrum rises to a peak and then falls on either side.

Step 2: Peak wavelength
\[ \lambda_m=\frac{2.89\times10^6\ \text{nm K}}{5780\ \text{K}}=500\ \text{nm} \]

Step 3: Compare
The wavelength 500 nm is exactly at the peak, so \(U_2\) is the largest. The wavelengths 300 nm and 900 nm are on opposite sides of the peak, so both energies are smaller than \(U_2\).

Step 4: Conclusion
\[ U_1<U_2>U_3 \]
This is option (C). We cannot say that \(U_1\) is more or less than \(U_3\) from the given data, and the answer does not need it.

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). \[ \boxed{U_1<U_2>U_3} \]