Question

Two spherical bodies a (radius 6 cm) and b(radius 18 cm) are at temperature 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?

Answer 1

Answer: The ratio of the total energy radiated is $\dfrac{P_{a}}{P_{b}} = \dfrac{9}{1}$

Explanation:

Given that,

Radius of a spherical body $r_{a} = 6 cm$

Radius of b spherical body $r_{b} = 18 cm$

Wave length of a $\lambda_{a} = 500 nm$

Wave length of b $\lambda_{b} = 1500 nm$

Using Wien's law

$\lambda_{A}\times T_{A} = \lambda_{B}\times T_{B}$...(I)

Now, put the value of $\lambda_{A}$ and $\lambda_{B}$ in equation (I)

$500\times T_{A} = 1500 \times T_{B}$

$\dfrac{T_{A}}{T_{B}} = \dfrac{3}{1}$

The power radiated is

$P = \sigma A T^{4}$

Therefore, the ratio of the total energy radiated

$\dfrac{P_{a}}{P_{b}} =\dfrac{\sigma A_{a} T^{4}_{a}}{\sigma A_{b}T^{4}_{b}}$

$\dfrac{P_{a}}{P_{b}} = \dfrac{r_{a}^{2}T_{a}^{4}}{r_{b}^{2}T_{b}^{4}}$

$\dfrac{P_{a}}{P_{b}} = (\dfrac{1}{3})^{2}\times (3)^{4}$

$\dfrac{P_{a}}{P_{b}} = \dfrac{9}{1}$

Hence, the ratio of the total energy radiated is $\dfrac{P_{a}}{P_{b}} = \dfrac{9}{1}$

Answer 2

Explanation:

answer of this question is 9.