Question

10 kg of ${U^{235}}$ is completely converted into energy. Then how much the energy will be released when?
A) $5 \times {10^{27}}MeV$
B) $5 \times {10^{24}}MeV$
C) $9 \times {10^{17}}J$
D) All of these

Answer 1

Hint: The basic approach that we are going to use in the solution is of
1) Mole concept which says that the number of particles in 1 mole of molecules is $6.022 \times {10^{23}}$ whose weight is equal to the atomic mass of the same molecule.
2) Secondly, we will be going to use the fixed value of energy in one atomic mass unit which is $931.5MeV$where the atomic mass unit is the average value of the isotopes of the same type in one atom.
3) Lastly, we will use the concept of unit conversion mainly for Mega that is ${10^6}$ and $1eV$ which is also called energy equals to charge on an electron magnitude.

Complete step by step answer:
According to the question given we have to find the mass of ten kilogram of uranium 235.
Let us start we the use of mole concept which that $235g$ of uranium contains 1 mole of molecules that is $6.022 \times {10^{23}}$
Now we will find the number of molecules in 10,000 grams. For this we will use the unitary method.
This method provides the number of different unit in one unit of the given substance, let us understand this mathematically,
Number of moles in one gram of Uranium $\dfrac{1}{{235}}mole$
Number of moles in 10,000 grams of Uranium $\dfrac{{10000}}{{235}} = 42.553moles$
We know that 1 mole contains $6.022 \times {10^{23}}$ atoms
So, 10,000 grams of Uranium contains $42.553 \times 6.022 \times {10^{23}} = 256.169 \times {10^{23}}$ Atoms
Now, Mass of 1 atom of uranium is $235amu = 235grams$ which contains energy of $931.5MeV$
So, $256.169 \times {10^{23}}$ Atoms has energy equals $256.169 \times {10^{23}} \times 931.5MeV \times 235 = 5.6 \times {10^{30}}MeV$
Converting further to $eV$ $,5.6 \times {10^{36}}eV$
$1eV = 1.6 \times {10^{ - 19}}Joules$
Hence the required value of energy after using the above conversion is
$8.96 \times {10^{17}}J = 9 \times {10^{17}}J$

Hence, the correct option is C

Note:
Now let us also look at another and easy way to solve this question is by using Einstein’s mass energy relationship that is ${E_{net}} = {m_T}{c^2}$ where c is the speed of light and m is the total mass
So, let us see the approach to solve the question by the Einstein’s method:
Value of speed of light is $c = 3 \times {10^8}m{s^{ - 2}}$
Mass is given $10kg$
So putting values in formula we get $9 \times {10^{17}}J$ as answer.