Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?

According to Wien's displacement law, λmT = constant, where λm is the wavelength of maximum intensity and T is the absolute temperature.

For body A, λm = 500 nm = 500 × 10-9 m. Let T1 be its temperature.
For body B, λm = 1500 nm = 1500 × 10-9 m. Let T2 be its temperature.

Using Wien's displacement law:
λmT = constant
500 × 10-9 T1 = 1500 × 10-9 T2
T1 = 3T2

According to Stefan-Boltzmann law, the rate of total energy radiated by a black body is proportional to the fourth power of its absolute temperature:
Rate of energy radiated ∝ T4

Let PA be the rate of total energy radiated by A and PB be the rate of total energy radiated by B.
Then,
PA ∝ T14
PB ∝ T24

The ratio of the rate of total energy radiated by A to that of B is:

PA / PB = T14 / T24 = (3T2)4 / T24 = 34 = 81

However, this solution considers only the temperature. The surface area also affects the total energy radiated. Let's include that. The surface area of sphere A is 4πrA² and the surface area of sphere B is 4πrB². The total power radiated is proportional to the surface area and T⁴.

PA ∝ (4πrA²)T1⁴ = (4π(6cm)²) (3T₂)^4
PB ∝ (4πrB²)T2⁴ = (4π(18cm)²) T₂⁴

PA/PB = [(4π(6cm)²) (3T₂)^4] / [(4π(18cm)²) T₂⁴] = (6²/18²)(3⁴) = (1/9)(81) = 9

Therefore, the ratio of the rate of total energy radiated by A to that of B is 9.