Question

The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Answer 1

Hint: Total energy, potential energy and kinetic energy are related to each other by the formula, K.E.= |T.E.|= $-\dfrac{\text{P}\text{.E}\text{.}}{2}$. Zero references of potential energy means the point where we are taking potential energy as zero.

Complete step by step answer:
(a) The kinetic energy of an electron is equal to the modulus or absolute value of total energy of an electron. Mathematically, represented as K.E. = |T.E.|. Modulus is taken because the kinetic energy of an electron is always positive. The absolute value of -3.4 eV will be 3.4 eV. The kinetic energy of an electron will be 3.4 eV.

(b) The potential energy of an electron is equal to twice the value of the total energy of an electron. Mathematically, represented as P.E. = $2\times $T.E. The potential energy is negative because the zero references are taken at the infinite energy level, so the energy of shells will be negative or less than zero. The kinetic energy of the electron will be $2\times $-3.4 eV or equal to -6.8 eV.

(c) The kinetic energy of an electron is not dependent on the zero references of potential energy. So, on changing the choice of zero references of potential energy, the kinetic will not change at all.
Potential energy will change depending on the choice of zero references. As we can define the reference point anywhere we want. Like, generally, the zero reference point is infinite but we can take potential energy at the first shell to be zero, then energies of the higher shells will be positive.

Note: The formula of potential energy is inversely proportional to distance from the nucleus. The formula of potential energy is $-\text{k}\dfrac{{{\text{e}}^{2}}}{\text{r}}$. The change in potential energy is $-\text{k}{{\text{e}}^{2}}\left( \dfrac{1}{{{\text{r}}_{1}}}-\dfrac{1}{{{\text{r}}_{2}}} \right)$. At an infinite point, the ${{\text{r}}_{2}}=\infty $, the potential energy is P.E. = $-\text{k}\dfrac{{{\text{e}}^{2}}}{{{\text{r}}_{1}}}$.