Question

In the hydrogen atom, the energy difference between the states n=2 and n=3 is Eev. The ionization potential of H atom is:
A.3.2 E
B.5.6 E
C.7.2 E
D.13.2 E

Answer 1

Hint: Before solving this question, we should first know the formula of Ionization potential. ${{E}_{n}}=-13.6ev\times \dfrac{{{Z}^{2}}}{{{n}^{2}}}$, we can put the given values in the given formula and can easily solve the question. Ionization energy can be defined as the estimation of the difficulty to eliminate an electron from an atom or the capability of an atom or ion to give an electron. It happens in the ground state of the atoms.

Complete answer:
The outermost electrons will face a less effective nuclear charge compared to the actual nuclear charge. It happens because the electrons that are innermost shield the electrons which are outermost by causing hindrance in the path of nuclear charge. For Example- If we consider an example of Sodium (Na), Inner electrons ($1{{s}^{2}},2{{s}^{2}},2{{p}^{6}}$) will shield the outer electron ($3{{s}^{1}}$).
 In the first case :
${{E}_{n}}=-13.6ev\times \dfrac{{{Z}^{2}}}{{{n}^{2}}}$
For hydrogen, Z = 1
${{E}_{n\,=\,2}}=-13.6ev\times \dfrac{{{1}^{2}}}{{{2}^{2}}}$
${{E}_{n\,=\,2}}=-13.6ev\times \dfrac{1}{4}$
${{E}_{n\,=\,3}}=-13.6ev\times \dfrac{{{1}^{2}}}{{{3}^{2}}}$
${{E}_{n\,=\,3}}=-13.6ev\times \dfrac{1}{9}$
${{E}_{n\,=\,3}}>{{E}_{n\,=\,2}}$
$E={{E}_{n\,=\,3}}-{{E}_{n\,=\,2}}$
$E=-13.6ev\times \dfrac{1}{9}+13.6ev\times \dfrac{1}{4}$
$E=\dfrac{5}{36}\times 13.6ev$
$13.6ev=\dfrac{36}{5}E$
In the second case :
${{E}_{n\,=\,\infty }}=-\dfrac{13.6ev\times {{1}^{2}}}{{{\infty }^{2}}}=0ev$
\[{{E}_{n\,=\,1}}=-\dfrac{13.6ev\times {{1}^{2}}}{{{1}^{2}}}=-13.6ev\]
${{E}_{n\,=\,\infty }}>{{E}_{n\,=\,1}}$
$I.P={{E}_{n\,=\,\infty }}-{{E}_{n\,=\,1}}$
$0+13.6ev=13.6ev$
We know $13.6ev=\dfrac{36}{5}E$
So, I.P = $\dfrac{36}{5}E$
      I.P = 7.2 E

So, Option (C) 7.2 E is correct.

Note:
To eliminate an electron from an atom needs energy in huge amounts to conquer the magnetic pull created by the positive charge of the nucleus. The I.E is always positive, As we move from left to right across a period, the ionization energy will always increase and as we move from top to the bottom, the ionization energy will decrease because the atomic number down the group increases results in the increase in the number of shells as well. The distance between the outermost electron and nucleus is so much that it can be easily removed. And as shells are increasing, the shielding effect will be more as well.