Question

The emissivities of the surfaces of two spheres P and Q of radii \(2R\) and \(3R\) are 0.35 and 0.7 respectively. The ratio of the powers radiated by the spheres P and Q is \(9:8\). If the wavelength at which sphere P emits radiations of maximum intensity is \(4000\ \text{\AA}\), then the wavelength at which sphere Q emits radiations of maximum intensity is:

• A. \(5000\ \text{\AA}\)
• B. \(3000\ \text{\AA}\)
• C. \(4500\ \text{\AA}\)
• D. \(6000\ \text{\AA}\)

Hint:

Use Stefan's law to find temperature ratio and then apply Wien's displacement law.

Answer 1

The Correct Option is D

Concept: According to Stefan's law, \[ P=e\sigma A T^4 \] and according to Wien's displacement law, \[ \lambda_m T=\text{constant}. \]

Step 1: Write power ratio.
\[ \frac{P_P}{P_Q} = \frac{e_P(4\pi (2R)^2)T_P^4} {e_Q(4\pi (3R)^2)T_Q^4}. \] \[ \frac{9}{8} = \frac{0.35\times4}{0.7\times9} \left(\frac{T_P}{T_Q}\right)^4. \] \[ \frac{9}{8} = \frac{2}{9} \left(\frac{T_P}{T_Q}\right)^4. \] \[ \left(\frac{T_P}{T_Q}\right)^4 = \frac{81}{16}. \] \[ \frac{T_P}{T_Q} = \frac32. \]

Step 2: Apply Wien's law.
\[ \lambda_P T_P=\lambda_Q T_Q. \] \[ \lambda_Q = \lambda_P\frac{T_P}{T_Q}. \] \[ = 4000\times\frac32. \] \[ =6000\ \text{\AA}. \] Hence \[ \boxed{6000\ \text{\AA}} \]