Solveeit Logo
Login

Question

Question: The ionization energy of the ground state of a hydrogen atom is \(2.18 \times {10^{ - 18}}{{J}}\). T...

The ionization energy of the ground state of a hydrogen atom is 2.18×10−18J2.18 \times {10^{ - 18}}{{J}}. The energy of an electron in its second orbit would be:
A. −1.09×10−18J - 1.09 \times {10^{ - 18}}{{J}}
B. −2.18×10−18J - 2.18 \times {10^{ - 18}}{{J}}
C. −4.36×10−18J - 4.36 \times {10^{ - 18}}{{J}}
D. −5.45×10−19J - 5.45 \times {10^{ - 19}}{{J}}

Explanation

Solution

Ionization energy is something which is used to measure the ability to remove the electron from an atom or ion. Generally, the elements in ground state lose electrons. It can also be defined as the strength by which electrons are placed in the orbits.

Complete step by step answer:
Ionization energy is a measure of the strength of bonds between atoms in a molecule. And also, it explains about reactivity. Usually ionization energy is expressed in electron volts, eV{{eV}} or kilojoules per mole, kJ.mol−1{{kJ}}{{.mo}}{{{l}}^{ - 1}}. Ionization energy is calculated by Bohr energy equation given below:
Eion=E∞−En⇔Eion=−13.6Z2n2{{{E}}_{{{ion}}}} = {{{E}}_\infty } - {{{E}}_{{n}}} \Leftrightarrow {{{E}}_{{{ion}}}} = - 13.6\dfrac{{{{{Z}}^2}}}{{{{{n}}^2}}}, where Eion{{{E}}_{{{ion}}}} is the ionization energy, E∞{{{E}}_\infty } is the energy at infinity level, En{{{E}}_{{n}}} is the energy at nth{{{n}}^{{{th}}}} level, Z{{Z}} is the atomic number and n{{n}} is the orbit number.
It is given that the ionization energy of the ground state of a hydrogen atom is 2.18×10−18J2.18 \times {10^{ - 18}}{{J}}. We know that the ionization energy is equal to 13.6eV{{13}}{{.6eV}}.
i.e. the ionization energy of first energy level, Eion=−13.6eV{{{E}}_{{{ion}}}} = - 13.6{{eV}}
Now we have to calculate the ionization energy of the electron in the second orbit using the above formula.
The atomic number of hydrogen is one, i.e. Z=1{{Z = 1}} and the orbit number, n=2{{n = 2}}.
Since 2.18×10−18J=−13.6eV2.18 \times {10^{ - 18}}{{J = - 13}}{{.6eV}}
i.e. E2=−2.18×10−18J1222=−2.18×10−18J4=−5.45×10−19J{{{E}}_{{2}}} = - 2.18 \times {10^{ - 18}}{{J}}\dfrac{{{1^2}}}{{{2^2}}} = \dfrac{{ - 2.18 \times {{10}^{ - 18}}{{J}}}}{4} = - 5.45 \times {10^{ - 19}}{{J}}
Thus the energy of electron in second orbit is −5.45×10−19J - 5.45 \times {10^{ - 19}}{{J}}

Hence, option D is the correct answer.

Additional information:
There are different classifications of ionization energies like first, second, third and so on. Second ionization energy is the energy to knock off the second electron from the atom or ion.

Note: Ionization energy is used to understand the bonds between metal molecules. Also, when we take the difference in the ionization energy, we can determine whether the bond is ionic or covalent. Ionization energy makes the atom or ion more positive.