Question

Mass equivalent to the energy 931 MeV is:
(A) $6.02 \times {10^{ - 27}}kg$
(B) $1.662 \times {10^{ - 27}}kg$
(C) $16.66 \times {10^{ - 27}}kg$
(D) $16.02 \times {10^{ - 27}}kg$

Answer 1

Hint: The mass-energy relationship can be given by the equation

\[E = m{c^2}\]

Where c = velocity of light in vacuum
Also \[1eV = 1.6 \times {10^{ - 19}}J\] .

Complete step by step answer:
Here, we are given the energy in the MeV unit and we are being asked to find its mass equivalent. So, we will first convert the energy in joules and then use the equation that relates mass to the energy.
- eV stands for electron volts and it is related with SI unit joule as follows. We know that charge on one electron is $1.6 \times {10^{ - 19}}C$. So,

\[1eV = 1.6 \times {10^{ - 19}}J\]

And M in MeV stands for mega and we can say that 1000eV = 1MeV. So, we can say that
931MeV = $931 \times 1.6 \times {10^{ - 19}} \times {10^6}J = 1489.6 \times {10^{ - 13}}J$
Now, we have the energy in the joule unit which is the SI unit. So, now we can use the mass-energy relation equation.
- Scientist Albert Einstein gave the relation between the mass and the energy. The formula is:

\[E = m{c^2}...........(1)\]

Where E = Energy
m = mass and c = velocity of the light in vacuum = $3 \times {10^8}m{s^{ - 1}}$
- We know the energy and velocity of light in vacuum. So, we can find the mass equivalent to that energy. So, putting available values in the equation (1), we get

\[1489.6 \times {10^{ - 13}} = m \times {(3 \times {10^8})^2}\]

So, we can write that

\[m = \dfrac{{1489.6 \times {{10}^{ - 13}}}}{{9 \times {{10}^{16}}}} = 165.51 \times {10^{ - 29}}kg = 1.6 \times {10^{ - 27kg}}\]

Thus, we can conclude that mass equivalent to the energy of 931MeV is $1.6 \times {10^{ - 27}}kg$.
So, the correct answer is (B).

Note: Remember that here we are using energy in the SI unit in the mass-energy relationship equation. So, the resultant mass obtained as a result will be in kg as it is the SI unit of mass. Remember that velocity of light in vacuum is $3 \times {10^8}m{s^{ - 1}}$, note here that the velocity value is in its SI unit form.