When a hydrogen atom is raised from the ground state to the excited state
- potential energy increases and K.E. decreases.
- potential energy decreases and K.E. increases.
- both K.E. and potential energy will increase.
- both K.E. and potential energy decreases.
Solution
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 $\text {P.E. }=-\frac{k Z e^2}{r}$
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. $\propto-\frac{1}{r}$
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 *
Asked in: MHT CET 2024 (09 May Shift 1)
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