Question

Fill in the blanks.
1 eV = ______________ Joule.

Answer 1

Hint: Electron volt is a measure of energy. It is defined by the energy lost or gained by an electron accelerating from rest through an electric potential difference of one volt in a vacuum. So, we can measure the gain in kinetic energy which is given by the multiplication of charge of electrons and the potential difference.

Complete step-by-step answer:
Joule is the SI unit of energy. In other terms,
1 Joule = $1kg-m/{{s}^{2}}$

As we can see, Joule is a very large unit and we cannot conveniently use it to deal with atomic or subatomic particles. So, we use the electron volt as the unit of energy in those cases. It is a very small unit. It is defined in the following way -

The electron volt is the amount of energy gained by an electron if it is accelerated through a vacuum with the help of one-volt potential difference.

We know that work done to move a charge of q through a potential difference of V is given by,
$KE=qV$

The charge of electrons is the most fundamental unit of charge. It is the lowest possible charge.

So, we have defined the electron volt in the mentioned manner. It is the work done to move the fundamental unit charge through a unit potential difference.

Charge of an electron is given by,
$q=1.602\times {{10}^{-19}}C$

So, 1 eV is given by,
$1eV=(1.602\times {{10}^{-19}}C)(1V)=1.602\times {{10}^{-19}}J$

Hence,
$1eV=1.602\times {{10}^{-19}}$ Joules

Note: The electron volt is used for multiple purposes. It can be used as Mass, momentum, distance, temperature, time, and energy. For example, you can convert 1 eV into the mass in the following way:

Using mass-energy equivalence, we can say that mass and energy are interchangeable. The unit of mass in terms of eV is given by,
$1\dfrac{eV}{{{c}^{2}}}=1.783\times {{10}^{-36}}kg$
The following diagram shows the uses of Electron Volt-

Measurement	Unit	SI value of unit
Energy	$eV$	$1.602\times {{10}^{-19}}J$
Mass	$\dfrac{eV}{{{c}^{2}}}$	$1.783\times {{10}^{-36}}Kg$
Momentum	$\dfrac{eV}{c}$	$5.344\times {{10}^{-28}}kg-m/s$
Temperature	$\dfrac{eV}{{{k}_{B}}}$	$1.160\times {{10}^{4}}K$
Time	$\dfrac{h}{eV}$	$6.582\times {{10}^{-16}}s$
Distance	$\dfrac{hc}{eV}$	$1.973\times {{10}^{-7}}m$