The ratio of ground state energy of Li^{2+},H | vedclass

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(C) The ground state energy of a hydrogen-like atom is given by the formula $E = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.Since all atoms are in the g

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$9:4:1$

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(C) The ground state energy of a hydrogen-like atom is given by the formula $E = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.Since all atoms are in the ground state,$n = 1$.Therefore,the energy is directly proportional to the square of the atomic number,$E \propto Z^2$.The atomic numbers are $Z = 3$ for $Li^{2+}$,$Z = 2$ for $He^{+}$,and $Z = 1$ for $H$.Thus,the ratio of their ground state energies is $E_{Li^{2+}} : E_{He^{+}} : E_{H} = (3)^2 : (2)^2 : (1)^2 = 9 : 4 : 1$.

List-$I$	List-$II$
$(1)$ Lyman	$(A)$ $n = 2 \to n = 1$
$(2)$ Balmer	$(B)$ $n = 3 \to n = 2$
$(3)$ Paschen	$(C)$ $n = 4 \to n = 3$
$(4)$ Brackett	$(D)$ $n = 5 \to n = 4$

The ratio of ground state energy of $Li^{2+}$,$He^{+}$,and $H$ is:

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List-$I$ List-$II$
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