Question

If the absolute temperature of a black body is doubled, the frequency corresponding to maximum emissive power becomes:

• A. halved
• B. doubled
• C. four times
• D. unchanged

Hint:

Remember: Temperature up $\rightarrow$ Wavelength down $\rightarrow$ Frequency up.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Wien's Displacement Law describes how the wavelength of maximum emission (\(\lambda_{max}\)) from a black body shifts with temperature. Frequency (\(\nu\)) and wavelength are inversely related.
Step 2: Key Formula or Approach:
Wien's Law: \(\lambda_{max} \cdot T = b\) (constant).
Relation between frequency and wavelength: \(c = \nu \cdot \lambda \implies \lambda = \frac{c}{\nu}\).
Step 3: Detailed Explanation:
Substitute \(\lambda = c/\nu\) into Wien's law:
\[ \frac{c}{\nu_{max}} \cdot T = b \]
\[ \nu_{max} = \frac{c \cdot T}{b} \]
Since \(c\) and \(b\) are constants, \(\nu_{max} \propto T\).
This means that the frequency at which the emissive power is maximum is directly proportional to the absolute temperature.

If temperature \(T\) is doubled:
\[ \nu_{max}' \propto 2T \implies \nu_{max}' = 2 \cdot \nu_{max} \]
Therefore, the frequency also becomes doubled.
Step 4: Final Answer:
The frequency is doubled.