Question

The mass of electron in MeV is:
                   c represents the speed of light.
A. $1.02MeV/{{c}^{2}}$
B. $0.51MeV/{{c}^{2}}$
C. \[51MeV/{{c}^{2}}\]
D. \[102MeV/{{c}^{2}}\]

Answer 1

Hint: According to the energy formula proposed by Albert Einstein energy of photons stored by the electrons is equal to the product of the mass to the square of its velocity. Obtained mass which is in kg can be converted in terms of MeV/c$^2$.
As per the given data,
C represents the speed of light that is $3\times {{10}^{8}}m{{s}^{-1}}$

Formula used:
$E=m{{c}^{2}}$

Complete step by step answer:
From the standard values,
We know that, the value of 1eV (a unit of energy) is equivalent to $1.60217657\times {{10}^{-19}}kg{{m}^{2}}{{s}^{-2}}$
According to the famous formula formulated by Albert Einstein in 1905. Energy equals mass times the square of the speed by which it is accelerating.
Mathematically.
$E=m{{c}^{2}}$
By putting the values:
$\begin{align}
  & 1.60217657\times {{10}^{-19}}=m{{(3\times {{10}^{8}})}^{2}} \\
 & m=\dfrac{1.60217657\times {{10}^{-19}}}{{{\left( 3\times {{10}^{8}} \right)}^{2}}} \\
 & m=1.780191888\times {{10}^{-36}}kg \\
\end{align}$
The rest mass of an electron is given by$9.1093891\times {{10}^{-31}}kg$.
The electron rest mass is the mass of a stationary electron. Rest mass is also termed as an invariant mass of an electron. This term is used because in special relativity cases the mass of an object can be said to get increased in a frame of reference with respect to the stationary or moving reference point.
So the rest mass of electrons constituting an energy of 1eV in terms of eV and c, will be given by,
$\begin{align}
  & {{m}_{e}}=\dfrac{9.10938291\times {{10}^{-31}}}{1.7801961888\times {{10}^{-36}}} \\
 & {{m}_{e}}=5.117066853226\times {{10}^{5}}eV/{{c}^{2}} \\
 & =0.51MeV/{{c}^{2}} \\
\end{align}$
Thus, the correct option which shows the exact value of mass of electrons in MeV is option B that is $0.51MeV/{{c}^{2}}$.

Note:
All the options are provided in terms of eV/c$^2$, so the answer has to be calculated in those terms only with using all the standard values properly (i.e. value of e and eV). The decimal places should be marked correctly for zero error.