Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature and , 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

Question

Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature $T_{1}$ and $T_{2}$ , 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? [2010]

Smart Achievers · Accepted Answer

(9)

$According to Stefan-Boltzmann law,$

$\frac{Rate of total energy radiated by A}{Rate of total energy radiated by B}$

$= \frac{σ T_{1}^{4} (4 π r_{1}^{2})}{σ T_{2}^{4} (4 π r_{2}^{2})} = {(\frac{T_{1}}{T_{2}})}^{4} \times {(\frac{r_{1}}{r_{2}})}^{2}$

$= {(\frac{λ_{m 2}}{λ_{m 1}})}^{4} {(\frac{r_{1}}{r_{2}})}^{2}$ $[∵ \frac{T_{1}}{T_{2}} = \frac{λ_{m_{2}}}{λ_{m_{1}}} by Wien's law]$

$= {(\frac{1500}{500})}^{4} {(\frac{6}{18})}^{2} = 9$

Solution

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Solution

Ans. (9) According to Stefan-Boltzmann law, Rate of total energy radiated by ARate of total energy radiated by B =σT14(4πr12)σT24(4πr22)=(T1T2)4×(r1r2)2 =(λm2λm1)4(r1r2)2 [∵T1T2=λm2λm1 by Wien's law] =(1500500)4(618)2=9

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