Question

Energy required for the electron excitation in $L{i^{ + + }}$ from the first to the third Bohr orbit is?
A. 12.1 eV
B. 36.3 eV
C. 108.8 eV
D. 122.4 eV

Answer 1

Hint: In order to give solution of the above question, we need to use the energy of ${n^{th}}$orbit of hydrogen atom so that we have to solve for the first orbit and the third orbit then we need to find the difference of two.

Complete step by step answer:
The energy of the electron in the ${n^{th}}$ orbit of a hydrogen atom is given by
${E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}eV$
For $L{i^{ + + }}$, Z=3
In the case of an first orbit, n = 1, where n be the number of orbital level
$
  {E_1} = \dfrac{{ - 13.6 \times {3^2}}}{1} \\
 \Rightarrow {E_1} = - 122.4eV \\
$
In the case of an third orbit, n=3
$
  {E_3} = \dfrac{{ - 13.6 \times {3^2}}}{{{3^2}}} \\
\Rightarrow {E_3} = - 13.6eV \\
$
Hence, the energy difference which is given by
$
  \Delta E = {E_2} - {E_1} \\
 \Rightarrow \Delta E = - 13.6 - ( - 122.4) \\
 \Rightarrow \Delta E = 108.8eV \\
$
Hence the correct option is C.

Note: Practically the electron which is present in a hydrogen atom carries certain energies. Such energies are usually called the energy levels of hydrogen. The quantum number n is denoted as the different energy levels of the hydrogen atom, where n varies from one to infinity. The first energy level is taken as the lowest energy level or ground state and the infinity is taken as the highest one.