Solveeit Logo
Login

Question

Question: How much energy is involved in ionizing hydrogen atoms at n=1 to 3?...

How much energy is involved in ionizing hydrogen atoms at n=1 to 3?

Explanation

Solution

Considering an Hydrogen atom, an electron in the Bohr’s Model is the most stable at its ground state i.e. when n =1 and the most unstable when it is at the infinite level i.e. n=∞n = \infty . The energy of electrons in the Bohr’s model is directly proportional to the square of the atomic number (Z) and inversely proportional to the square of the principal quantum number (n).

Complete answer:
The energy of the electron in a orbit according to Bohr model can be given as:
E=−13.6z2n2E = - 13.6\dfrac{{{z^2}}}{{{n^2}}}
Where Z is the atomic number and n is the no. of orbits present in the atom.
We are given that initially the electron is present in n=1 (Ground state). The energy of the ground state will be; (The atomic number of hydrogen is 1)
E1=−13.61212=−13.6eV{E_1} = - 13.6\dfrac{{{1^2}}}{{{1^2}}} = - 13.6eV
The excited state given to us is n = 3. The energy of the electron in that state is: E2=−13.61232=−1.511eV{E_2} = - 13.6\dfrac{{{1^2}}}{{{3^2}}} = - 1.511eV
The energy required for the excitation will be the difference between the two: ΔE=E2−E1\Delta E = {E_2} - {E_1}
ΔE=−1.511−(−13.6)\Delta E = - 1.511 - ( - 13.6)
ΔE=−1.511+13.6\Delta E = - 1.511 + 13.6
ΔE=+12.10eV\Delta E = + 12.10eV
To convert eV in Joules we’ll use the converting factor 1eV=1.6×10−19J1eV = 1.6 \times {10^{ - 19}}J
12.10eV=12.10×1.6×10−19J=1.938×10−18J12.10eV = 12.10 \times 1.6 \times {10^{ - 19}}J = 1.938 \times {10^{ - 18}}J
This is the energy required to excite the electron from n=1 to n=3. Now If we have to find individual energies of every orbital n=1, n=2, n=3, we can find it using the Rydberg’s equation.
E1=−13.612eV=−13.6eV(n=1){E_1} = \dfrac{{ - 13.6}}{{{1^2}}}eV = - 13.6eV(n = 1)
E2=−13.622eV=−3.40eV(n=2){E_2} = \dfrac{{ - 13.6}}{{{2^2}}}eV = - 3.40eV(n = 2)
E3=−13.632eV=−1.51eV(n=3){E_3} = \dfrac{{ - 13.6}}{{{3^2}}}eV = - 1.51eV(n = 3)
According to Koopman’s approximation the ionization energies of an orbital containing one electron is of the approximately same magnitude as the energy of that orbital. Therefore, the ionization energies will be: E1=−13.6eV,E2=−3.40eV,E3=−1.51eV{E_1} = - 13.6eV,{E_2} = - 3.40eV,{E_3} = - 1.51eV

Note:
Ionization energies are always negative, because a huge amount of energy is released when an electron is lost, the reaction is exothermic. At infinity, the energy of the atom is zero as there are no electrostatic forces acting on it.