Question

The energy of an electron in the first energy level is $ - 21.79 \times {10^{ - 12}} $ erg per atom. The energy of an electron in the second energy levels is:
(A) $ - 54.47 \times {10^{ - 12}}\;erg\,ato{m^{ - 1}} $
(B) $ - 5.447 \times {10^{ - 12}}\;erg\,ato{m^{ - 1}} $
(C) $ - 0.5447 \times {10^{ - 12}}\;erg\,ato{m^{ - 1}} $
(D) $ - 0.05447 \times {10^{ - 12}}\;erg\,ato{m^{ - 1}} $

Answer 1

Hint: Energy levels are the set of discrete values of the total energy of subatomic particles confined in the system. Energy levels are also called the energy states. The first energy level is nearest to the nucleus. The second energy level is a little away from the first.
Formula used:
The energy of an electron in $ {n^{th}} $ level, $ {E_n} = \dfrac{{13.6{Z^2}}}{{{n^2}}}erg\;at{m^{ - 1}} $
Where $ {E_n} $ is the energy of an electron in the $ {n^{th}} $ level, $ n $ is the orbital or level, and $ Z $ is the atomic number.

Complete step by step answer:
Now we have the idea of the energy of the electron and the energy level. So, now with the help of basic mathematics, we can easily solve the question. To solve any numerical first we used to write the given quantities. So, according to the question, the energy of an electron in the first energy level is $ - 21.79 \times {10^{ - 12}} $ erg per atom. So for the first energy level, we know that $ n = 1 $ now we can represent it as $ {E_1} = - 21.79 \times {10^{ - 12}}\;erg\,ato{m^{ - 1}} $ . So, now we will calculate the energy of an electron in the second energy level $ {E_2} $ .
Now from the formula of the energy of an electron for a particular element, $ {E_n}\alpha \dfrac{1}{{{n^2}}} $ . So, we can write the energy of an electron for the two levels as $ \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{\left( {n_2^2} \right)}}{{\left( {n_1^2} \right)}} $ . Now for the first energy level we have $ {E_1} = - 21.79 \times {10^{ - 12}}\;erg\,ato{m^{ - 1}},{n_1} = 1 $ and for the energy of an electron in the second energy level we have $ {E_2},{n_2} = 2 $ . Now we will substitute these values in the above formula we get,
 $ \Rightarrow \dfrac{{ - 21.79 \times {{10}^{ - 12}}}}{{{E_2}}} = \dfrac{{{{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^2}}} $
After solving we get,
 $ {E_2} = - 5.447 \times {10^{ - 12}}{\text{erg}}\;{\text{at}}{{\text{m}}^{ - 1}} $ .
Therefore, the correct option is (B).

Note:
It is to be noted that the electrons at the higher energy levels have more energy as compared to the electrons at lower energy levels. Electrons at a higher energy level will have more orbitals and the possible number of electrons.