Question

The total energy of an electron in the first excited state of the hydrogen atom is $-3.4eV$. Find out its (i) kinetic energy and (ii) potential energy in this state.

Answer 1

Hint: We know that the law of conservation of energy states that the energy can neither be created nor it can be destroyed. But it can only be converted from one form to another. So, in case of an atom, we need to find the total energy as the sum of the kinetic energy and the potential energy. This principle states that in a system that does not undergo any force from the outside of the system, the amount of the energy is constant, irrespective of its changes in the form. Based on this concept we have to solve this question.

Complete step by step answer:
At first let us find the kinetic energy of the electron:
Kinetic energy of electron is,
$KE=\dfrac{13.6{{Z}^{2}}}{{{n}^{2}}}eV$
In the above expression Z denotes the atomic number and n denotes the shell number at which the electron is present inside the atom.
For the first excited state of the hydrogen atom, n = 2 and Z = 1
$\therefore KE=\dfrac{13.6}{{{2}^{2}}}=3.4eV$
Total energy as we know that will be given as:
$E=KE+PE\Rightarrow -3.4=3.4+PE$
E denotes the total amount of energy in the above expression, KE is the kinetic energy and the PE denotes the potential energy in the above expression.
$\therefore PE=-6.8eV$
Hence, the answer is $-6.8eV$ .
So, at the end we can say that:

The value of the kinetic energy is 3.4 eV
The value of the potential energy is -6.8 eV

Note: We know that Niels Bohr proposed a theory which is popularly known as the Bohr’s atomic theory. The theory states that for the hydrogen atom based on the quantum theory that energy is transferred only in certain well-defined quantities. We should also have an idea that electrons should move around the nucleus but only in the orbits that are prescribed orbits. However, when jumping from one orbit to another with lower energy, a light quantum is emitted.