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Question

(A) For an electron in a hydrogen-like atom,the relationship between kinetic energy $(KE)$ and potential energy $(PE)$ is given by $PE = -2 	imes KE$

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$+\frac{3}{2}x$

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(A) For an electron in a hydrogen-like atom,the relationship between kinetic energy $(KE)$ and potential energy $(PE)$ is given by $PE = -2 	imes KE$.Initial state: $KE_{1} = x$,so $PE_{1} = -2x$.Final state: $KE_{2} = \frac{x}{4}$,so $PE_{2} = -2 	imes (\frac{x}{4}) = -\frac{x}{2}$.The change in potential energy is $\Delta PE = PE_{2} - PE_{1}$.$\Delta PE = -\frac{x}{2} - (-2x) = -\frac{x}{2} + 2x = +\frac{3x}{2}$.

An electron in an atom jumps in such a way that its kinetic energy changes from $x$ to $\frac{x}{4}$. The change in potential energy will be

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