Question

A body P at 1000 K emits maximum energy at a wavelength of 3000 nm. If another body Q emits maximum energy at wavelength 550 nm, what will be the temperature of that body Q?

• A. 5454 K
• B. 6250 K
• C. 3125 K
• D. 4000 K

Hint:

Wien's Law (\(\lambda_{\text{max}} T = \text{constant}\)) is perfect for ratio problems involving the peak emission wavelength and temperature of black bodies. Remember that as an object gets hotter, its peak emission shifts to shorter wavelengths (bluer color).

Answer 1

The Correct Option is A

Step 1: Wien's Displacement Law: The wavelength of maximum radiation emission (\( \lambda_{\text{max}} \)) from a black body is inversely proportional to its absolute temperature (\(T\)). Mathematically:\[ \lambda_{\text{max}} T = b \]where \(b\) is Wien's displacement constant (\( \approx 2.898 \times 10^{-3} \) m⋅K).

Step 2: Ratio Setup for Bodies P and Q: Because \( \lambda_{\text{max}} T \) is constant:\[ \lambda_{P} T_{P} = \lambda_{Q} T_{Q} \]

Step 3: Solve for Unknown Temperature \(T_Q\): Rearrange the formula:\[ T_Q = T_P \left( \frac{\lambda_P}{\lambda_Q} \right) \]

Step 4: Calculate \(T_Q\): Substitute values:- \( T_P = 1000 \) K- \( \lambda_P = 3000 \) nm- \( \lambda_Q = 550 \) nmNote: Nanometer units can be used in the ratio without conversion. Calculation:\[ T_Q = 1000 \times \frac{3000}{550} = 1000 \times \frac{300}{55} = 1000 \times \frac{60}{11} \approx 1000 \times 5.4545 \]\[ T_Q \approx 5454.5 \text{ K} \]