Question

The ionisation potential of hydrogen atom is
$\text{A}\text{. }12.97V$
$\text{B}\text{. }10.2V$
$\text{C}\text{. }13.6V$
$\text{D}\text{. }27.2V$

Answer 1

Hint: The ionisation energy of an atom is given by ${{E}_{n}}=-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV$. The ionisation energy per unit charge is called ionisation potential. Use the formula Z$V=13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}V$ to find the ionisation potential of hydrogen atom.
Formula used:
${{E}_{n}}=-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV$
$V=\dfrac{E}{e}$
$V=13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}V$

Complete step by step answer:
Atoms consist of many imaginary shells or orbits in which the electrons of the atoms revolve. The shells are denoted by K, L, M ,N and so on. Since the revolving electrons are bound to the nucleus or are being attracted to the nucleus due to the electromagnetic force, the electrons possess some energy. The energy of an electron consists of some potential energy and some kinetic energy.
The energy of an electron depends on which orbits it is revolving. Therefore, these orbits are also called energy levels, denoted as ${{n}^{th}}$ energy level. The first orbit that is the K shell is the first energy level (i.e. n = 1).
The energy of an electron present in ${{n}^{th}}$ level is equal to ${{E}_{n}}=-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV$.
Here Z is the atomic number of the atom.
When the outermost electron of an atom is taken from its respective ${{n}^{th}}$ orbit to an orbit that is at infinity, the atom is said to be ionised. And this process is called ionisation.
To make the electron jump from ${{n}^{th}}$ orbit to infinity, we have to impart a specific amount of energy to the electron. The energy that has to be given to the electron is called ionisation energy and is equal to $E={{E}_{\infty }}-{{E}_{n}}$.
Therefore,
$E={{E}_{\infty }}-{{E}_{n}}=-13.6\dfrac{{{Z}^{2}}}{{{\infty }^{2}}}eV-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV=0-\left( -13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV \right)$
$\Rightarrow E=13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV$.
The ionisation energy per unit charge is called ionisation potential (V). Here, it the charge of the electron. Therefore, the ionisation potential is equal to $V=\dfrac{E}{e}$.
This implies,
 $V=\dfrac{13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV}{e}=13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}V$.
For a hydrogen atom Z = 1 and it has only one electron in the first orbit i.e. n = 1.
Therefore, ionisation potential of hydrogen atom is
  $V=13.6\dfrac{{{1}^{2}}}{{{1}^{2}}}V=13.6V$
Hence, the correct option is C.

Note: The given formula for the energy of an electron in ${{n}^{th}}$ orbit i.e. ${{E}_{n}}=-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}}eV$ is only applicable for mono electronic species like hydrogen atom and ions like helium.
If the atom has more than one electron then the formula changes to ${{E}_{n}}=-13.6\dfrac{{{\left( Z-b \right)}^{2}}}{{{n}^{2}}}eV$, where b is called screening constant.
The value of b depends on the repulsions due to the other electrons present in the atom.