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: Understanding the Question:
The question asks how the wavelength of maximum emission ($\lambda_{max}$) from a black body changes if its absolute temperature ($T$) is doubled.
This is governed by the laws of thermal radiation.

Step 2: Key Formula or Approach:
We apply Wien's Displacement Law, which relates the peak wavelength of black body radiation to its absolute temperature:
\[ \lambda_{max} \cdot T = b \]
Where $b$ is Wien's displacement constant ($b \approx 2.898 \times 10^{-3}\text{ m}\cdot\text{K}$).
This indicates that:
\[ \lambda_{max} \propto \frac{1}{T} \]

Step 3: Detailed Explanation:

• Wien's Displacement Law states that the wavelength ($\lambda_{max}$) at which the emissive power of a black body is maximum is inversely proportional to its absolute temperature ($T$).

• Let the initial absolute temperature be $T_1 = T$ and the initial wavelength of maximum emission be $\lambda_1$.

• The final temperature is doubled: $T_2 = 2T$.

• Let the final wavelength of maximum emission be $\lambda_2$.

• Using the inverse proportionality:
\[ \frac{\lambda_2}{\lambda_1} = \frac{T_1}{T_2} \]

• Substitute the values into the ratio:
\[ \frac{\lambda_2}{\lambda_1} = \frac{T}{2T} = \frac{1}{2} \]
\[ \lambda_2 = \frac{\lambda_1}{2} \]

• Thus, doubling the absolute temperature causes the wavelength of maximum emission to be halved.


Step 4: Final Answer:
The wavelength of maximum emission is halved.