Question

The kinetic energy of electron in the first Bohr orbit will be:
A. 13.6 eV
B. -489.6 eV
C. 0.38 eV
D. -0.38 eV

Answer 1

Hint: Kinetic energy of any particles is not negative, so negative options can be ruled out. Hydrogen has an electron revolving in the first Bohr's orbit so derivation can be done by only considering the motion of the electron in hydrogen ground state.

Formula used:
Total energy of an electron in n${}^{th}$ orbit of hydrogen is given as:
$E_n={\displaystyle \frac{-13.6}{n^2}}$eV.

Complete step by step answer:
For an electron the total energy can be written as sum of kinetic energy due to its motion and potential energy due to the Coulomb attraction.
$E={\displaystyle \frac{1}{2}}m{v}^2-{\displaystyle \frac{K{e}^2}{r}}$ ;
where K is a constant factor and since we are considering hydrogen, we get only positive charge of e in the nucleus (balancing e of a single electron in its orbit).
As the electron is undergoing a circular motion around the nucleus we may write:
${\displaystyle \frac{m{v}^2}{r}}={\displaystyle \frac{K{e}^2}{r^2}}$,
where we simply equated centripetal force with coulomb force.
We can use this for substitution in kinetic energy expression, we will get:
$K.E.={\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{K{e}^2}{2r}}$ ;
which will give us the total energy as:
$E={\displaystyle \frac{K{e}^2}{2r}}-{\displaystyle \frac{K{e}^2}{r}}$
$E=-{\displaystyle \frac{K{e}^2}{2r}}$
But we also know that the total energy is given by
$E_n={\displaystyle \frac{-13.6}{n^2}}$eV
which becomes -13.6eV for n =1.
Therefore, we may write:
$-13.6eV=-{\displaystyle \frac{K{e}^2}{2r}}$
Substituting this value in the kinetic energy expression that we obtained above, we get:
$K.E.={\displaystyle \frac{K{e}^2}{2r}}=13.6eV$ .
Therefore, the kinetic energy for an electron in first Bohr's orbit is 13.6eV. So, the correct answer is option (A).

Note:
The total energy for an electron is negative but kinetic energy will always be positive. For higher orbits, the total energy will decrease as n will increase. This energy is also called ionization energy; it signifies that electrons in higher orbits are not as much strongly attracted by the potential of the nucleus and can be easily separated from the atom.