Question

If the PE of the ground state electrons is taken as zero then find the total energy of the first excited stage.

Answer 1

Hint:In this question we are asked to calculate the total energy of the first excited stage if the potential energy of the ground state electrons is taken as zero. To solve this question, we will the total energy of the electrons in the ground and first excited stage. We will then calculate the potential energy of electrons in ground state. Later, we will assume the PE in ground state to be zero and thus by further calculation find the total energy of the first excited stage.
Formula used:
\[E=\dfrac{-13.6}{{{n}^{2}}}\]
Where,
E is the energy
n is the number of energy levels or the principal quantum number

Complete answer:
We know that energy of any level in a hydrogen atom is given by,
\[E=\dfrac{-13.6}{{{n}^{2}}}\]
Now, the energy for ground level i.e. n = 1 will be,
E = -13.6 eV …………… (by substituting n=1 in equation) …………… (1)
Similarly, the energy for first excited stage i.e. n = 2 will be,
\[E=\dfrac{-13.6}{{{2}^{2}}}\]
E = -3.4 eV ………………. (2)
Now, we know that potential energy of the ground state is -27.2 eV. Therefore, if we are to add +27.2 eV to it the potential energy will become zero. Similarly, if we add +27.2 eV to the first excited state, we will get the energy of the first excited state when potential energy is zero.
Now adding +27.2 to equation (2)
We get,
\[-3.4+27.2\]
= 23.2 eV
Therefore, if the PE of the ground state electron is assumed zero, the TE of the first excited state will be 23.8 eV.

Note:
In quantum mechanics, the principal quantum number is one of the four quantum numbers assigned to an electron. It determines the state the electron. 1 eV is the energy an electron gains when it travels through a potential difference of Volt. When an electron absorbs energy it is said to be in an excited state. This excited state may be unstable, therefore, it returns to the next lowest stable energy level by releasing energy.