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Byju's Answer

Standard XII

Chemistry

Lanthanides Electronic Configuration

A Hydrogen at...

Question

A Hydrogen atom and a $L i^{+ +}$ ion are both in the second excited state. If $l_{H}$ and $l_{L i}$ are their respective electronic angular momenta, and $E_{H}$ and $E_{L i}$ their respective energies, then

l_{H} > l_{L i}

and

| E_{H} | > | E_{L i} |

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l_{H} = l_{L i}

and

| E_{H} | < | E_{L i} |

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l_{H} = l_{L i}

and

| E_{H} | > | E_{L i} |

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l_{H} < l_{L i}

and

| E_{H} | < | E_{L i} |

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Solution

The correct option is B $l_{H} = l_{L i}$ and $| E_{H} | < | E_{L i} |$
As we know,

$l = \frac{n h}{2 π} & | E | \propto \frac{Z^{2}}{n^{2}}$

Here, $n = 2; Z_{H} = 1; Z_{L i} = 3$

$∴ l_{H} = l_{L i}$ and $| E_{H} | < | E_{L i} |$

Hence, option $(B)$ is correct.

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A Hydrogen atom and a Li++ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then

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Here, $n = 2; Z_{H} = 1; Z_{L i} = 3$

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Hence, option $(B)$ is correct.