Question

The power obtained in a reactor using ${{\text{U}}^{235}}$ disintegration is $1000{\text{kW}}$ . The mass decay of ${{\text{U}}^{235}}$ per hour is
(A) $10$ microgram
(B) $20$ microgram
(C) $40$ microgram
(D) $1$ microgram

Answer 1

Hint : To solve this question, we first have to calculate the energy obtained in the reactor in the given time from the value of the power given. Then, we have to use Einstein's mass energy equation to get the value of the mass decay corresponding to this value of the energy.

Formula used: The formulae which are used to solve this question are given by
$\Rightarrow E = Pt$ , here $E$ is the energy obtained in time $t$ due to a power of $P$ .
$\Rightarrow E = m{c^2}$ , here $E$ is the energy corresponding to a mass of $m$ , and $c$ is the value of the speed of light in the vacuum.

Complete step by step answer
According to the question, the power obtained through the disintegration of the ${{\text{U}}^{235}}$ nucleus is equal to $1000{\text{kW}}$ . Therefore we have
$\Rightarrow P = 1000{\text{kW}}$
Since, $1{\text{kW}} = 1000{\text{W}}$ . So the power is
$\Rightarrow P = 1000 \times 1000{\text{W}}$
$\Rightarrow P = {10^6}{\text{W}}$ ………………………………….(1)
Now, the energy is related to the power by
$\Rightarrow E = Pt$ ………………………………….(2)
According to the question, we have the time
$\Rightarrow t = 1h$
We know that $1h = 3600s$ . So we have
$\Rightarrow t = 3600s$ ………………………………….(3)
Substituting (1) and (3) in (2) we get
$\Rightarrow E = {10^6} \times 3600{\text{J}}$
$\Rightarrow E = 3.6 \times {10^9}{\text{J}}$ ………………………………….(4)
Now, from the Einstein’s mass energy equation we have
$\Rightarrow E = m{c^2}$ ………………………………….(5)
The speed of light in vacuum is
$\Rightarrow c = 3 \times {10^8}m/s$ ………………………………….(6)
Substituting (4) and (6) in (5) we have
$\Rightarrow 3.6 \times {10^9} = m{\left( {3 \times {{10}^8}} \right)^2}$
$\Rightarrow m = \dfrac{{3.6 \times {{10}^9}}}{{9 \times {{10}^{16}}}}kg$
On solving we get
$\Rightarrow m = 4 \times {10^{ - 8}}kg$
Since, $1kg = 1000g$ . So we have
$\Rightarrow m = 4 \times {10^{ - 8}} \times 1000g$
$\Rightarrow m = 4 \times {10^{ - 5}}g$
As, $1g = {10^6}\mu g$ . So we finally get
$\Rightarrow m = 4 \times {10^{ - 5}} \times {10^6}\mu g$
$\Rightarrow m = 40\mu g$
Hence, the mass decay of ${{\text{U}}^{235}}$ per hour is equal to $40$ micrograms.
Hence, the correct answer is option C.

Note
We need to be careful regarding the units of the quantities. We should remember to convert all the values in their respective SI units. Also, in this question, the name of the nucleus is given just to confuse us. There is no need at all to use this information in solving this question.