Question

The energy of the electron in the first Bohr’s orbit of the H atom is -13.6 eV. What will be its potential energy in n = 4^th orbit?
(a) -13.6 eV
(b) -3.4 eV
(c) -0.85 eV
(d) -1.70 eV

Answer 1

Hint: We know that Bohr's theory was proposed for the hydrogen atom which says that the movement of electrons around the nucleus takes place only in specific orbits. And when the electrons jump to an orbit of lower energy, emission of radiation takes place.

Formula used: The formula for the energy possessed by an electron in the nth orbit is,
\[{E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV\]. Here, n represents the orbit number.

Complete Step by Step Answer:
Here, first, we will find the electron's energy in the 4th orbit of the atom of hydrogen.
\[{E_n} = \dfrac{{ - 13.6}}{{{4^2}}}eV = - 0.85\,eV\]

So, the total energy in the 4th orbit is -0.85 eV.

Now, we have to find the relation between total energy and potential energy. In the atom of hydrogen, potential energy is two times the total energy in the nth orbit of the electron.
So, the relation is,

\[{E_P} = 2{E_n}\]
Here, \[{E_P}\] stands for potential energy, and \[{E_n}\] stands for energy at the nth orbit.
We got, \[{E_n} = - 0.85\,eV\]

Therefore,
\[{E_P} = 2{E_n}\]
\[{E_P} = 2 \times - 0.85 = - 1.70\,{\rm{eV}}\]
Therefore, the electron's potential energy is -1.70 eV.
Hence, the correct option is (D).

Note: It is to be noted that, the total energy of an electron in higher orbits decreases with the increase of the number of orbits. It is because of the fact that the electrons of the higher energy orbits are not strongly attracted by the nucleus. Thus, the removal of the electrons is easy.