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: State Wien's Displacement Law. Wien's law states that the wavelength at which a black body emits the maximum amount of radiation, \( \lambda_{\text{max}} \), is inversely proportional to its absolute temperature \(T\). \[ \lambda_{\text{max}} T = b \] where \(b\) is Wien's displacement constant (\( \approx 2.898 \times 10^{-3} \) m⋅K).

Step 2: Set up a ratio for the two bodies P and Q. Since \( \lambda_{\text{max}} T \) is constant for any black body: \[ \lambda_{P} T_{P} = \lambda_{Q} T_{Q} \]

Step 3: Rearrange the formula to solve for the unknown temperature, \(T_Q\). \[ T_Q = T_P \left( \frac{\lambda_P}{\lambda_Q} \right) \]

Step 4: Substitute the given values and calculate \(T_Q\). - \( T_P = 1000 \) K - \( \lambda_P = 3000 \) nm - \( \lambda_Q = 550 \) nm Note that since we are using a ratio, we do not need to convert nanometers to meters. \[ T_Q = 1000 \times \frac{3000}{550} = 1000 \times \frac{300}{55} = 1000 \times \frac{60}{11} \approx 1000 \times 5.4545 \] \[ T_Q \approx 5454.5 \text{ K} \]