Question

What is the rest energy of electrons?
A. $$510\text{KeV}$$
B. $$931\text{KeV}$$
C. $$510\text{MeV}$$
D. $$931\text{MeV}$$

Answer 1

Hint: Use the formula, Einstein’s mass energy equivalence, $$E=m{c}^2$$. By knowing the values, substitute in the following equation. To find the energy in electron volt, divide the energy in joules by $$\text{1}\text{.602}\times \text{1}{\text{0}}^{-\text{19}}\text{J}$$.

Complete step by step solution:
The rest mass of an electron is the mass of a stationary electron, also known as the electron's invariant mass. This is one of physics' fundamental constants. Often the term "rest mass" is used because in special relativity an object's mass can be said to increase in a frame of reference that moves relative to that object (or if the object moves within a given frame of reference). Many practical experiments on the travelling electrons are carried out.

The rest energy of any material can be related by the formula as postulated by Einstein’s mass-energy equivalence, which is given below
$$E=m{c}^2$$ …… (1)
Here,
$$E$$ indicates the energy of the electron.
$$m$$indicates the mass of the electron.
$$c$$indicates the velocity of the electron at which the electron moves which may be considered as the speed of light.

So, by substituting the value of $$1\text{eV = 1}\text{.602}\times \text{1}{\text{0}}^{-\text{19}}\text{J}$$and $$c=3\times {10}^8\text{m}{\text{s}}^{-\text{1}}$$ in equation (1):
$$\begin{gathered} E=m{c}^2 \\ =9.1\times {10}^{-31}\text{kg}\times {\left(3\times {10}^8\text{m}{\text{s}}^{-\text{1}}\right)}^2 \\ =9.1\times {10}^{-31}\times 3^2\times {\left({10}^8\right)}^2\text{J} \\ =9.1\times 9\times {10}^{-31}\times {10}^{16}\text{J} \\ =81.3\times {10}^{-15}\text{J} \end{gathered}$$
The energy found above is in joules, but it needs to be converted into electron volt.
We know:
$$1\text{eV = 1}\text{.602}\times \text{1}{\text{0}}^{-\text{19}}\text{J}$$
So, to convert the energy into electron volt, divide the energy in joules by $$\text{1}\text{.602}\times \text{1}{\text{0}}^{-\text{19}}\text{J}$$.
$$\begin{gathered} E={\displaystyle \frac{81.3\times {10}^{-15}\text{J}}{\text{1}\text{.602}\times \text{1}{\text{0}}^{-\text{19}}\text{Je}{\text{V}}^{-\text{1}}}} \\ =50.74\times {10}^{-15+19}\text{eV} \\ \text{ = }50.74\times {10}^4\text{eV} \\ \text{ = 507}\text{.4}\times {10}^3\text{eV} \\ \text{ = 507}\text{.4 KeV} \end{gathered}$$

The energy found is $$\text{507}\text{.4 KeV}$$ which can be rounded off as $$\text{510 KeV}$$.

Hence, the rest energy of electron is $$\text{510 KeV}$$.

Note: In the above problem, you are required to find the rest energy of the electron. To do this just apply Einstein's mass energy equivalence $$E=m{c}^2$$. While doing the problem, always take the speed of light in $$\text{m}{\text{s}}^{-\text{1}}$$ but not in $$\text{km}{\text{h}}^{-\text{1}}$$. If you take $$\text{km}{\text{h}}^{-\text{1}}$$, it will definitely affect the result.