An electron in a hydrogen-like atom undergoes a transition in such a way that its kinetic energy cha — Atomic Structure Chemistry Question
Question
An electron in a hydrogen-like atom undergoes a transition in such a way that its kinetic energy changes from $K$ to $\frac{K}{4}$. What is the corresponding exact change in its potential energy ($\Delta PE$)?
Answer: A
💡 Solution & Explanation
In any Bohr orbit, the potential energy $PE = -2KE$. Initial state: $PE_1 = -2K$. Final state: $KE_2 = \frac{K}{4}$, so $PE_2 = -2(\frac{K}{4}) = -\frac{K}{2}$. The change in potential energy $\Delta PE = PE_2 - PE_1 = (-\frac{K}{2}) - (-2K) = 2K - \frac{K}{2} = +\frac{3K}{2}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes