The black discs x, y and z have radii 1 m, 2 m and 3 m respectively. The wavelengths corresponding to maximum intensity are 200 nm, 300 nm and 400 nm respectively. The relation between emissive power Ex, Ey and Ez is (a) Ex > Ey > Ez (b) Ex Ex < Ez

Answer: (a). Power E ∝ R² T⁴ ∝ (R²)/(λm⁴). Taking λ in units of 100 nm: Ex ∝ 1/2⁴ = 0.0625, Ey ∝ 4/3⁴ = 0.049, Ez ∝ 9/4⁴ = 0.035. So Ex > Ey > Ez, answer (a).

The black discs x, y and z have radii 1 m, 2 m and 3 m respectively… | MHT CET 2024 (10 May, Shift 2)

Q69MHT CET 2024 · 10 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien

The black discs $x, y$ and z have radii $1\text{ m}, 2\text{ m}$ and 3 m respectively. The wavelengths corresponding to maximum intensity are $200\text{ nm}, 300\text{ nm}$ and 400 nm respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is

Show answer and solution

Answer: (a)

Power $E \propto R^2 T^4 \propto \frac{R^2}{\lambda_m^4}$ .
Taking $\lambda$ in units of 100 nm: $E_x \propto \frac{1}{2^4} = 0.0625$ , $E_y \propto \frac{4}{3^4} = 0.049$ , $E_z \propto \frac{9}{4^4} = 0.035$ .
So $E_x > E_y > E_z$ , answer (a).

Report a mistake

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2024 (10 May, Shift 2)

More questions on Thermal Radiation: Black Body, Stefan & Wien