In Bohr's model of hydrogen-like species,whic | vedclass

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(A) The Kinetic Energy $(KE)$ of an electron in a hydrogen-like species is given by the formula: $KE = 13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For hyd

Vedclass · Accepted Answer

$KE$ of electron in ground state of hydrogen atom is equal to $KE$ of electron in $n = 3$ state of $Li^{2+}$

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(A) The Kinetic Energy $(KE)$ of an electron in a hydrogen-like species is given by the formula: $KE = 13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For hydrogen $(H)$,$Z = 1$. For $Li^{2+}$,$Z = 3$.Checking option $A$: $KE$ of $H$ in ground state $(n=1)$ is $13.6 	imes \frac{1^2}{1^2} = 13.6 	ext{ eV}$. $KE$ of $Li^{2+}$ in $n=3$ is $13.6 	imes \frac{3^2}{3^2} = 13.6 	ext{ eV}$.Since both values are equal,option $A$ is correct.Checking option $B$: $KE$ of $H$ in $n=3$ is $13.6 	imes \frac{1^2}{3^2} = 1.51 	ext{ eV}$. $KE$ of $Li^{2+}$ in ground state $(n=1)$ is $13.6 	imes \frac{3^2}{1^2} = 122.4 	ext{ eV}$. These are not equal.Checking option $C$: Total energy $E = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$. Energy of $H$ in ground state $(n=1)$ is $-13.6 	ext{ eV}$. Energy of $Li^{2+}$ in $n=2$ is $-13.6 	imes \frac{3^2}{2^2} = -30.6 	ext{ eV}$. These are not equal.

In Bohr's model of hydrogen-like species,which of the following is true?

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