Question

When a hydrogen atom is raised from ground state to excited state.
A) Both K.E. and P.E. increases.
B) Both K.E. and P.E. decreases.
C) The P.E. increases and K.E. decreases.
D) The K.E. decreases and P.E. increases.

Answer 1

Here a hydrogen atom is raised from ground to excited state while doing so its energy will change. We have to find whether its potential and kinetic energy both increases or decreases or one of them increases while the other decreases. Hence we can use the formula of P.E. and K.E. of hydrogen atom to solve this problem.
Formula used:

& K.E.=\dfrac{kZ{{e}^{2}}}{2r} \\\ & P.E.=-\dfrac{kZ{{e}^{2}}}{r} \\\ \end{aligned}$$ **Complete answer:** Here we have asked about the change in potential energy and kinetic energy of the hydrogen atom when it is raised from ground to excited level. Now the formula for kinetic energy for hydrogen atom is given as $$K.E.=\dfrac{kZ{{e}^{2}}}{2r}$$ Where k is constant, e is charge of electrons revolving around the nucleus, r is the radius of the orbit or distance of the nucleus and the electron revolving around it and Z is the number of protons in the nucleus. Now when an electron is excited to a higher energy level it means that it has occupied a higher orbital and as the orbital increases the distance of electron from the nucleus increases i.e. r increases. And from the above formula we can write $$K.E.\propto \dfrac{1}{r}$$ That is kinetic energy is inversely proportional to the radius of orbit. If r increases then K.E. decreases. Hence we can conclude that when an electron is excited to a higher energy level its kinetic energy decreases. Similarly formula for potential energy for the hydrogen atom can be given as $$P.E.=-\dfrac{kZ{{e}^{2}}}{r}$$ Where k is constant, e is charge on an electron, r is radius of orbital and Z is the number of protons in the nucleus. We can see that potential energy is also inversely proportional to the radius of orbital and we can write it as $$P.E.\propto -\dfrac{1}{r}$$ Although it is inversely proportional but due to negative sign, when the radius is increased, the potential energy is also increased. Hence, when a hydrogen atom is raised from ground state to excited state, its potential energy increases whereas kinetic energy decreases. **So, the correct answer is “Option C”.** **Note:** Here one can make the mistake that potential energy also decreases as it is inversely proportional to radius of orbital but if we calculate it numerically due to negative sign when radius increases potential energy also increases. The ground state energy represents the energy required to free an electron from its shell or orbit.