91Ó°ÊÓ

Here, in this problem, we need to determine the mass converted into radiation in a year by a quasar. The given data is as follows

The rate at which the quasar radiates energy is$R={10}^{41}W={10}^{41}J/s$

One solar mass unit is$1smu=2.0×{10}^{30}kg$ .

Quasars were called Quasi-Stellar objects because most of the radiation is emitted by the nucleus of the active galaxy, which looks like a star. Active galaxy nucleus emits radiation at the order of 10⁴⁰J/S

The quasar, which is the nuclei of a galaxy, emits radiation of energy 10⁴¹ in one second. Using the energy equivalence principle to determine the mass converted in this process in one second.

$$\begin{gathered} E=\mathrm{Δ}m{c}^2 \\ \mathrm{Δ}m=\frac{E}{c^2} \end{gathered}$$

Substitute all the value in the above equation

$$\begin{gathered} \mathrm{Δ}m=\frac{{10}^{41}J}{{\left(3×{10}^8\right)}^2} \\ =1.1×{10}^{24}kg \end{gathered}$$

So, in one second, $1.1×{10}^{24}kg$is converted into energy. Therefore, in one year, the amount of mass that is converted is

$\left(1.1×{10}^{24}Kg/s\right)\left(\frac{60×60×24×365s}{1y}\right)=3.5×{10}^{31}kg/y$

Expressing the result in solar mass units, we get,

$$\begin{gathered} \frac{3.5×{10}^{31}kg/y}{2.0×{10}^{30}kg/smu}=17.5smu/y \\ ~18smu/y \end{gathered}$$

Hence, the rate at which the mass is being reduced from a quasar is$18smu/y$$18smu/y$.