Question

The potential energy of an electron in the hydrogen atom is $- 6.8eV$. Indicate in which excited state the electron is present?
(A) First
(B) Second
(C) Third
(D) Fourth

Answer 1

For an electron revolving around the nucleus of an atom which has only one electron revolving around it, the total energy of the system is equal to the sum of the potential energy of the electron and the kinetic energy of the electron in that excited state.

Complete step by step answer:
We are given the potential energy of the electron in the hydrogen atom and required to find the excited state it is present in.
Bohr’s theory was based on the application of Planck’s quantum theory of the atomic spectra of hydrogen atoms.
The postulates of the theory are: -
The electron in an atom has only certain definite stationary states which are called as energy levels.in each of the energy levels the electron revolves in a circular motion, due to the centripetal force acting on the electron from the centre.
Only those states allow electronic motion which have the angular momentum of an electron as an integral multiple of $\dfrac{h}{{2\pi }}$.
Now as we know that the force on the electron is centripetal, therefore the electrostatic force between the electron and nucleus is
Let the charge on the electron be ‘e’ and that will be the same charge on the nucleus due to just one proton. Also, the mass of electron is defined as ${m_e}$
$\dfrac{{KZ{e^2}}}{{{r^2}}} = \dfrac{{{m_e}{v^2}}}{r}$ …. Eq (1)
Also, we know that only those states are allowed that have the angular momentum as an integral multiple of $\dfrac{h}{{2\pi }}$
Therefore
${m_e}vr = n\dfrac{h}{{2\pi }}$ …… Eq (2)
Solving equation one and two, we get
$r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}KZ}} \\\ \\\$
Putting the values of the above given quantities, we get
$n = 1 \\\ h = 6.626 \times {10^{ - 35}}Js \\\ K = 9 \times {10^9}N{m^2}{C^{ - 2}} \\\ e = 1.6 \times {10^{ - 19}}C \\\$
We get
$r = 0.529 \times {10^{ - 10}}\dfrac{{{n^2}}}{Z}$
The hydrogen atom is of hydrogen therefore atomic number is one and as we need to find out the first energy state therefore ‘n’ is also one.
$r = 0.529 \times {10^{ - 10}}m$
Now the energy can be defined as the energy needed to take out one electron completely out of protons reach, now for that energy can be written as
$E = - \dfrac{{KZ{e^2}}}{{2r}}$
Putting the values, we get
$E = - 2.178 \times {10^{ - 18}}J \times \dfrac{{{Z^2}}}{{{n^2}}}$
Dividing by the charge on an electron that is $1.6 \times {0^{ - 19\,}}C$we get the energy present here in the SI Unit electron volt
$E = - 2.178 \times {10^{ - 18}}J \times \dfrac{{{Z^2}}}{{{n^2}}}$
$\Rightarrow E = \dfrac{{ - 2.178 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}} \times \dfrac{{{Z^2}}}{{{n^2}}}eV$
$\Rightarrow E = - 13.6 \times \dfrac{{{Z^2}}}{{{n^2}}}eV$
We get the energy in a random excited state as $E = - 13.6 \times \dfrac{{{Z^2}}}{{{n^2}}}eV$
Now we are given that the potential energy of the electron in an excited state as the double of the total energy in terms of magnitude and also the sign of the total energy is also negative due to that
Therefore, if the potential energy is given as $- 6.8eV$ then the total energy would be the half of it and that would be
$\dfrac{{ - 6.8eV}}{2} \\\ = - 3.4eV \\\$
Now equating this with the formula $E = - 13.6 \times \dfrac{{{Z^2}}}{{{n^2}}}eV$ of total energy we get the excited state’s number
Here the value of atomic number is one because it is just an hydrogen atom
$E = - 13.6 \times \dfrac{{{Z^2}}}{{{n^2}}}eV = - 3.4eV$
$\Rightarrow - 13.6 \times \dfrac{{{1^2}}}{{{n^2}}}eV = - 3.4eV$
$\Rightarrow {n^2} = 4$
$\Rightarrow n = 2$
The value of n came out to be two, therefore the excited state would be the first excited state.
Therefore, the electron is in the first excited state. So, option (A) is correct.

Note:
For a Bohr atom the radius is directly proportional to the energy state and inversely proportional to the atomic number, while energy is directly proportional to the atomic number and inversely proportional to the energy state.