Question

Which is not a correct order of energy for 1st , 2nd , and 3rd orbit?
A.${E_1} > {E_2} > {E_3}$
B.${\left( {PE} \right)_1} < {\left( {PE} \right)_2} < {\left( {PE} \right)_3}$
C.${\left( {KE} \right)_1} > {\left( {KE} \right)_2} > {\left( {KE} \right)_3}$
D.A and C both

Answer 1

Hint:The energy of the electron rotating in an orbit in the atom is equal to the sum of potential energy of the electron as a result of the attraction between the nucleus and the electrons and the kinetic energy of the electrons due to their motion around the nucleus. The energy of orbit is inversely related to its radius.

Complete step by step solution:
The mathematical representation of the energy of the orbitals is as follows:
$E$=$\dfrac{{{E_0}}}{{{n^2}}}$
where “n” is the number of the shell, or the principal quantum number.
So the energy of the electron in an orbital is inversely proportional to the square of the principal quantum number. So the order of the energies should be: ${E_1} > {E_2} > {E_3}$ and the order of kinetic energies is also ${\left( {KE} \right)_1} > {\left( {KE} \right)_2} > {\left( {KE} \right)_3}$.
As both A and C are the correct options.

So, the correct answer is option D.

Note:
The ${E_0}$ is the above equation is the minimum energy required by the electron to stay in the nucleus and is often called the threshold energy of the electron. The value of this energy is $13.6{\text{eV}}$. The principal quantum number or the number of the shell is always a natural number. The electrons absorb or release energy only when it transfers from one shell to another otherwise it revolves around the nucleus in shells of fixed energies.