StreakPeaked· Practice

ExamsJEE AdvancedPhysics

Two objects X and Y have thermal emissivities 0.01 and 0.81 respectively. Their outer surface areas are equal. Both radiate energy at the same rate. The wavelength lambda_Y corresponding to the peak spectral radiance of Y is shifted from the peak wavelength of X by 1.00 micrometres. Given that the temperature of X is 5802 K, find the temperature of Y and its peak wavelength lambda_Y.

  1. The temperature of Y is 1934 K
  2. lambda_Y = 1.5 micrometres
  3. The temperature of Y is 11604 K
  4. The temperature of Y is 2901 K

Correct answer: The temperature of Y is 1934 K

Solution

Setting emissive power equal: 0.01 * (5802)⁴ = 0.81 * T_Y⁴, so T_Y⁴ = (0.01/0.81) * (5802)⁴ = (1/81) * (5802)⁴, giving T_Y = 5802/3 = 1934 K. By Wien's law, lambda_X = 2898/5802 ~ 0.5 micrometres and lambda_Y = 2898/1934 ~ 1.5 micrometres, so the shift = 1.5 - 0.5 = 1.0 micrometre, consistent with the given data. Options A and B are both correct.

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