# A quasar consumes 1 solar mass of material per year, converting 15 percent of it directly into energy. What is the quasar's luminosity, in solar units?

Feb 1, 2016

$L = 1.478 \times {10}^{13} {L}_{\odot}$

#### Explanation:

Quasars are created as a result of mas being swallowed by the supermassive black hole at the center of an active galactic nucleus. The luminosity of a quasar is the rate at which energy is emitted by the quasar. We are given the rate of $1 {M}_{\odot} \text{/ yr}$, and Einstein gave us the formula to convert mass into energy.

$E = m {c}^{2}$

Here, $m$ is the mass converted and $c$ is the speed of light. One solar mass is equal to $1.989 \times {10}^{30} \text{kg}$. Using the mass conversion formula;

$E = {\left(1.989 \times {10}^{30} \text{kg")(3 xx 10^8 "m/s}\right)}^{2}$

$E = 1.790 \times {10}^{47} \text{J}$

Luminosity is measured in watts, $\text{W}$, or $\text{J/s}$, so we need to convert years into seconds.

$1 \text{year" = 365"days" = 8,760 "hours" = 525,600 "minutes}$
$= 31 , 536 , 000 \text{seconds}$

So the luminosity of the quasar is;

L = (1.790 xx 10^47 "J")/(3.154 xx 10^7 "s") = 5.675 xx 10^39 "W"

For comparison, the luminosity of the Sun is;

${L}_{\odot} = 3.828 \times {10}^{26} \text{W}$

Which is about 1 ten-billionth of the luminosity of the quasar.

$L = \left(5.675 \times {10}^{39} \text{W")/(3.282 xx 10^26"W}\right) = 1.478 \times {10}^{13} {L}_{\odot}$