Solveeit Logo
Login

Question

Question: The energy of an electron in the first Bohr orbit of H-atom is \( - 13.6eV\). The possible energy va...

The energy of an electron in the first Bohr orbit of H-atom is −13.6eV - 13.6eV. The possible energy value of the electron in the excited state of Li2+L{i^{2 + }} is
A. −122.4eV - 122.4eV
B. 30.6eV30.6eV
C. −30.6eV - 30.6eV
D. 13.6eV13.6eV

Explanation

Solution

Here we have Li2+L{i^{2 + }}, which has a hydrogen-like structure. A hydrogen-like structure means that it has only one electron. We can use the energy equation to find out the solution.

Formula Used:
The energy of nth{n^{th}} the orbit of other hydrogen-like one electron is given as
En=−13.6n2Z2  eV/atom{E_n} = - \dfrac{{13.6}}{{{n^2}}}{Z^2}\;eV/atom
Where n is the nth{n^{th}} the energy level and Z is the atomic number.

Complete step by step solution:
Bohr’s atomic model was introduced to explain the atomic structure of hydrogen and hydrogen-like species. That means which has only one electron in its shell. For example, He+,  Li2+H{e^ + },\;L{i^{2 + }}etc.,
We know that the orbits in which electrons are revolving have a fixed value of energy. These orbits are called energy levels or stationary states. As long as electrons are revolving in these orbitals, they neither lose nor gain energy and hence explain the stability of the atom. The energy of hydrogen atom is given as
En=−2π2me4n2h2{E_n} = \dfrac{{ - 2{\pi ^2}m{e^4}}}{{{n^2}{h^2}}}
Where n is the nth{n^{th}} energy level.
Putting the values of mmand ee
En=−13.6n2eV/atom{E_n} = - \dfrac{{13.6}}{{{n^2}}}eV/atom
Therefore, the energy of the first shell where n=1n = 1
The energy of the shell is −13.6  eV/atom - 13.6\;eV/atom
The energy of nth{n^{th}} the orbit of other hydrogen-like one electron is given as
En=−13.6n2Z2  eV/atom{E_n} = - \dfrac{{13.6}}{{{n^2}}}{Z^2}\;eV/atom
Where n is the nth{n^{th}} the energy level and Z is the atomic number.

So to find the possible energy value of an electron in the excited state of Li2+L{i^{2 + }} we can use the above equation.
Substituting n=1n = 1 and Z=3Z = 3
E1=−13.61×32=−13.6×9{E_1} = - \dfrac{{13.6}}{1} \times {3^2} = - 13.6 \times 9
E1=−122.4eV{E_1} = - 122.4eV

**Therefore the energy of an electron in the excited state is E1=−122.4eV{E_1} = - 122.4eV.

Note:**
Apart from energy, we can also find the radius of Bohr’s orbit.
For nth{n^{th}} the orbit of any one electron species with atomic number Z is
rn=0.529Z×n2{r_n} = \dfrac{{0.529}}{Z} \times {n^2}
i.e., r∝n2r \propto {n^2}