The energy of an electron in first Bohr orbit of H-atom is $- 13.6 e V$ . The possible energy value of electron in the excited state of ${L i}^{2 +}$ is:

- 122.4 e V

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30.6 e V

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- 30.6 e V

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13.6 e V

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Solution

The correct option is B $- 30.6 e V$
$E_{n} = \frac{E_{1}}{n^{2}} \times Z^{2}$

where
$E_{1} =$ Energy of first Bohr orbit $= - 13.6 e V$
$n =$ no of orbit
$Z =$ Atomic no

For ${L i}^{2 +}$ , the excited state, $n = 2$ and $Z = 3$

$∴$ $E_{n} = \frac{- 13.6}{2^{2}} \times {(3)}^{2}$
$= \frac{- 13.6}{4} \times 9 = - 30.6 e V$

Hence, the correct option is $C$

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The energy of an electron in first Bohr orbit of H-atom is −13.6 eV. The possible energy value of electron in the excited state of Li2+ is:

The correct option is B −30.6 eVEn=E1n2×Z2where E1=Energy of first Bohr orbit =−13.6eVn= no of orbitZ=Atomic noFor Li2+, the excited state, n=2 and Z=3∴ En=−13.622×(3)2=−13.64×9=−30.6 eVHence, the correct option is C

The energy of an electron in first Bohr orbit of H-atom is $- 13.6 e V$ . The possible energy value of electron in the excited state of ${L i}^{2 +}$ is: