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
To solve this question we need to use the Bohr’s model of the hydrogen atom. From there we can easily derive the expression for the kinetic energy. And the potential energy is given by the potential at a point, where the potential at infinity is zero.
The formulas used to solve this problem is,
$L = mvr = n\hbar$ where $n = 1,2,3,...$ which is principal quantum number, $L$ is angular momentum of an electron in a hydrogen atom, $m$ is mass of the electron, $v$ is the velocity of the electron in the orbit, $r$ is the radius of the orbit
$\hbar = \dfrac{h}{{2\pi }} = 1.054 \times {10^{ - 34}}J/\sec$ where $h$ is Planck’s constant
$\dfrac{{m{v^2}}}{r} = \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _o}{r^2}}}$ where $Z$ is the atomic number which is 1 in the case of hydrogen

Complete Step by Step Solution
The angular momentum of an electron in a hydrogen atom is quantized by the formula,
$L = mvr = n\hbar$
This is the second postulate of the Bohr theory.
From the condition of revolution of the electron in the ${\text{nth}}$ orbit,
$\dfrac{{m{v^2}}}{r} = \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _o}{r^2}}}$ , which is Bohr’s first postulate.
Therefore, by multiplying $\dfrac{1}{2}$on both sides and cancelling the $r$on both sides we get,
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{Z{e^2}}}{{8\pi {\varepsilon _o}r}}$
Now we know the term on the L.H.S gives kinetic energy which we can denote by $T$. So,
$\Rightarrow T = \dfrac{1}{2}m{v^2} = \dfrac{{Z{e^2}}}{{8\pi {\varepsilon _o}r}}$
If we consider the potential at infinity to be zero, the electric potential of a point charge $e$ brought from infinity to a point at a distance $r$ away from a point charge $Ze$ is given by,
$\Rightarrow V = - \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _o}r}}$
Thus, this potential energy of an electron in the orbit of an atom where $Ze$ is the nuclear charge.
From the values of $T$ and $V$we can say that,
$\Rightarrow V = - \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _o}r}} = - 2T$
So the total energy of an electron is given by the sum of kinetic and potential energies,
$E = T + V$
$\Rightarrow E = T + \left( { - 2T} \right)$
Therefore, we get,
$\Rightarrow E = - T$
Hence, the kinetic energy is $- E = - \left( { - 3.4eV} \right) = 3.4eV$

So the answer to the first part (i) is $3.4eV$.

And the potential energy,
$\Rightarrow V = - 2T = - \left( {2 \times 3.4eV} \right) = - 6.8eV$

So the answer to the second part (ii) is $- 6.8eV$.

Note
We should take proper care of the negative and positive signs while deriving the formula for the total energy. From the Bohr’s first and second postulate we can also find the expressions for the radius of the orbit $r$and the velocity of the electron in that orbit $v$.