According to Bohr's model, the highest kineti | vedclass

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(B) The kinetic energy $(KE)$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom with atomic number $Z$ is given by the formula: $KE = 13.6

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First orbit of $He^{+}$

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(B) The kinetic energy $(KE)$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom with atomic number $Z$ is given by the formula: $KE = 13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.Calculating for each option:$(A)$ For the first orbit of $H$ atom $(Z=1, n=1)$: $KE = 13.6 	imes \frac{1^2}{1^2} = 13.6 	ext{ eV}$.$(B)$ For the first orbit of $He^{+}$ $(Z=2, n=1)$: $KE = 13.6 	imes \frac{2^2}{1^2} = 13.6 	imes 4 = 54.4 	ext{ eV}$.$(C)$ For the second orbit of $He^{+}$ $(Z=2, n=2)$: $KE = 13.6 	imes \frac{2^2}{2^2} = 13.6 	ext{ eV}$.$(D)$ For the second orbit of $Li^{2+}$ $(Z=3, n=2)$: $KE = 13.6 	imes \frac{3^2}{2^2} = 13.6 	imes 2.25 = 30.6 	ext{ eV}$.Comparing the values, the highest kinetic energy is $54.4 	ext{ eV}$, which corresponds to the first orbit of $He^{+}$.

According to Bohr's model, the highest kinetic energy is associated with the electron in the

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