Question

Total energy of the electron in hydrogen atom above 0eV leads to
A. continuation of energy states
B. large number of discrete ionized states
C. Palmer series
D. Paschen series

Answer 1

You could recall the total energy of an electron in a hydrogen atom. You may see that the energy has a negative sign as the indication of the electron being bound to the nucleus. So in the question we are asked about the total energy being above 0eV which clearly indicates that the electron will no longer be bound to the nucleus. Keeping this fact in mind, find the answer from the options.
Formula used:
Expression for Total energy,
$E=-\dfrac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}$

Complete answer:
As a first step let us derive the expression for total energy in a hydrogen atom.
From the Rutherford atom model that is based on classical concepts, the atom is pictured as an electrically neutral sphere that consists of an electrically neutral sphere that has minute but massive positive nucleus at the centre with electrons revolving around it in stable orbits. It is the electrostatic force of attraction between the nucleus and the electron that keeps the electron in the orbits by providing centripetal force. Therefore, in a stable orbit,
${{F}_{E}}={{F}_{C}}$
$\Rightarrow \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{e}^{2}}}{{{r}^{2}}}=\dfrac{m{{v}^{2}}}{r}$
$\Rightarrow {{v}^{2}}=\dfrac{r{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}m}=\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}rm}$
So the kinetic energy of an electron in a hydrogen atom is given by,
$K.E=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}m\left( \dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}rm} \right)$
$\Rightarrow K.E=\dfrac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}$ …………………………… (1)
Also, we know the electrostatic potential energy of electron in a hydrogen atom is given by,
$P.E=-\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}$ ……………………….. (2)
Also, the Total energy is given by the sum of (1) and (2),
$\Rightarrow T.E=K.E+P.E$
$\Rightarrow E=\dfrac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}+\left( -\dfrac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r} \right)$
$\Rightarrow E=-\dfrac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}$
We see that the total energy of the electron in a hydrogen atom is negative, which is a clear indication of the fact that the electron is bound to the nucleus and not free. If this energy was positive, then, the electron might not be bound to the nucleus and hence will be free.
So, we could say that, as long as the total energy is negative and bound to the nucleus, energy levels of the electron are quantized, that is, they can only have a definite energy. But when the total energy becomes positive, the electron is no more bound to the nucleus and is free. So, the energy levels will have continuous values.
Therefore, the total energy of the electron in a hydrogen atom above 0eV means the energy is positive, that means the electron’s energy states are continuous.

So, the correct answer is “Option A”.

Note:
Also if the total energy of the electron was positive, then the electron will no longer follow the closed path around the nucleus. The lowest energy state in an atom is the ground state. Total energy of this state has the value of $-13.6eV$.