A hydrogen atom and a $L i^{+ +}$ ion are both in second excited state. If $l_{H} a n d l_{L i}$ are their respective electronic angular momenta, and $E_{H}$ and $E_{L i}$ their respective energies, then
(IIT-JEE-2002)

$l_{H} > l_{L i} a n d | E_{H} | > | E_{L i} |$

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$l_{H} = l_{L i} a n d | E_{H} | < | E_{L i} |$

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$l_{H} = l_{L i} a n d | E_{H} | > | E_{L i} |$

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$l_{H} < l_{L i} a n d | E_{H} | < | E_{L i} |$

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Solution

The correct option is B
$l_{H} = l_{L i} a n d | E_{H} | < | E_{L i} |$

Since the angular momentum of an electron in the Bohr’s Atomic model is $l = \frac{n h}{2 π} .$ It only depends on the principle quantum number. So, $l_{H} = l_{L i} .$

Also, Since energy is directly proportional to the square of the atomic number. The magnitude of energy of Lithium ion is more than the magnitude of energy of Hydrogen atom.

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