Question

A star behaves like a perfectly black body emitting radiant energy. The ratio of radiant energy emitted per second by this star to that emitted by another star having 8 times the radius of former, but having temperature one-fourth of that of the former in kelvin is :
A. 1 : 4
B. 1 : 16
C. 4 : 1
D. 16 : 1

Answer 1

Hint: A black body emits radiant energy in the form of electromagnetic waves of all the possible wavelengths. Use the formula for the rate of radiant energy emitted by a perfectly black body and then find the ratio of the two. Use the given relation between the radii and the temperatures of the two stars.
Formula used:
$E=\sigma A{{T}^{4}}$

Complete answer:
It is given that the star behaves as a perfectly black body. A perfectly black body is that body that absorbs all the radiations that are incident on it. Hence, a perfectly black body absorbs all the radiant energy. Also, a black body emits radiant energy in the form of electromagnetic waves of all the possible wavelengths.
The rate of energy emitted (energy emitted per second) by a perfectly black body at a temperature T is given as $E=\sigma A{{T}^{4}}$,
where $\sigma $ is the Stefan’s constant and A is the surface area of the body.
Let the radius of the first star be ${{r}_{1}}$.
$\Rightarrow {{A}_{1}}=4\pi r_{1}^{2}$.
Let the temperature of this star be ${{T}_{1}}$
 Therefore, the energy emitted per second by the first star is ${{E}_{1}}=\sigma {{A}_{1}}T_{1}^{4}=4\pi \sigma r_{1}^{2}T_{1}^{4}$.
Let the radius of the second star be ${{r}_{2}}$.
$\Rightarrow {{A}_{2}}=4\pi r_{2}^{2}$.
Let the temperature of this star be ${{T}_{2}}$
Since bodies are stars, the Stefan constant will be the same for both.
Therefore, the energy emitted per second by the second star is ${{E}_{2}}=\sigma {{A}_{2}}T_{2}^{4}=4\pi \sigma r_{2}^{2}T_{2}^{4}$.
Now,
$\Rightarrow \dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{4\pi \sigma r_{1}^{2}T_{1}^{4}}{4\pi \sigma r_{2}^{2}T_{2}^{4}}$
$\Rightarrow \dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{r_{1}^{2}T_{1}^{4}}{r_{2}^{2}T_{2}^{4}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}{{\left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}$ ….. (i).
According to the given data we get that $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{1}{8}$ and $\dfrac{{{T}_{1}}}{{{T}_{2}}}=4$.
Substitute these values in equation (i).
 $\Rightarrow \dfrac{{{E}_{1}}}{{{E}_{2}}}={{\left( \dfrac{1}{8} \right)}^{2}}{{\left( 4 \right)}^{4}}=4$.
The ratio of radiant energy emitted by the first start to the second is 4 : 1.

So, the correct answer is “Option C”.

Note:
The formula for the radiant energy per second that we used in the solution is only applicable for a perfectly black body.
The actual formula for the rate of emission of radiant energy is given as $E=\sigma eA{{T}^{4}}$, where e is the emissivity of the black body. The emissivity of a perfectly black body is equal to 1 and it is the maximum value.
For other black bodies, e < 1.