Question

If $13.6eV$ energy is required to ionize the hydrogen atom, then the energy required to remove an electron from its excited state in orbit $n = 2$ is equal to
(A) $10.2eV$
(B) $0eV$
(C) $3.4eV$
(D) $6.8eV$

Answer 1

Hint : The hydrogen is the element with only one proton and one electron. This electron will excite to higher energy levels when the energy is supplied and returns back to ground state after some time. The energy required to remove the electron from its excited state can be determined from the below formula.
${E_n} = \dfrac{{13.6}}{{{n^2}}}$
${E_n}$ is the energy required to remove an electron from excited state
n is the number of orbits.

Complete Step By Step Answer:
Ionization is the phenomenon of making a molecule into ions. There are two types of ions. They are positive ions and negative ions. The ion with positive charge can be called as cation and the ion with negative charge can be called as an anion.
Hydrogen is the element with atomic number $1$ . The number of protons in hydrogen is $1$ and the number of neutrons is $1$ . When this hydrogen atom is ionized, it produces an electron and proton. Given that $13.6eV$ energy is required to ionize the hydrogen atom.
The energy required to remove an electron from its excited state in orbit $n = 2$ will be obtained by substituting the value of $n = 2$ in the above formula
$E = \dfrac{{13.6}}{{{2^2}}} = \dfrac{{13.6}}{4} = 3.4eV$
Thus, the energy is $3.4eV$
Option C is the correct one.

Note :
This concept was introduced by scientist Bohr and was successful in determining the energy required to remove an electron from the atoms involving one proton like helium cation and lithium cation. The energy can easily be calculated by knowing the number of orbits to which the electron is excited.