Question

If the temperature of a black body is doubled, the wavelength of maximum emission:

• A. Doubles
• B. Halves
• C. Becomes 4 times
• D. Remains same

Hint:

Wien's Law states $\lambda_{max} \propto 1/T$.
As a black body gets hotter, it shifts its peak emission to shorter wavelengths (higher frequencies), which is why heating metal makes it change color from red to yellow and then blue.

Answer 1

The Correct Option is B

Step 1: Recall Wien's displacement law.
The wavelength at which a black body emits most strongly is inversely related to its absolute temperature:
\[ \lambda_{max} T = b \text{ (a constant)} \]
Step 2: Apply the inverse relationship.
Since \( \lambda_{max} = b/T \), doubling the temperature means the new peak wavelength is \( b/(2T) \).
Step 3: Compare to the original.
\[ \frac{\lambda_{max}'}{\lambda_{max}} = \frac{b/(2T)}{b/T} = \frac{1}{2} \]
This is exactly why very hot objects like stars glow blue-white (shorter wavelength) while cooler ones glow red (longer wavelength).
\[ \boxed{\text{Halves}} \]