Energy and radius of first Bohr orbit of He + and Li 2 + are [Given R H = 2 . 18 × 10 - 18 J , a 0 = 52 . 9 pm ] [2025]

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Solution

Q.
Energy and radius of first Bohr orbit of ${He}^{+}$ and ${Li}^{2 +}$ are [Given $R_{H} = 2.18 \times 10^{- 18} J,$ $a_{0} = 52.9$ $pm$ ] [2025]

1 $E_{n} ({Li}^{2 +}) = - 19.62 \times 10^{- 16} J; r_{n} ({Li}^{2 +}) = 17.6 pm$ $E_{n} ({He}^{+}) = - 8.72 \times 10^{- 16} J; r_{n} ({He}^{+}) = 26.4 pm$

2 $E_{n} ({Li}^{2 +}) = - 8.72 \times 10^{- 16} J; r_{n} ({Li}^{2 +}) = 17.6 pm$ $E_{n} ({He}^{+}) = - 19.62 \times 10^{- 16} J; r_{n} ({He}^{+}) = 17.6 pm$

3 $E_{n} ({Li}^{2 +}) = - 19.62 \times 10^{- 18} J; r_{n} ({Li}^{2 +}) = 17.6 pm$ $E_{n} ({He}^{+}) = - 8.72 \times 10^{- 18} J; r_{n} ({He}^{+}) = 26.4 pm$

4 $E_{n} ({Li}^{2 +}) = - 8.72 \times 10^{- 18} J; r_{n} ({Li}^{2 +}) = 26.4 pm$ $E_{n} ({He}^{+}) = - 19.62 \times 10^{- 18} J; r_{n} ({He}^{+}) = 17.6 pm$

Ans.
(3)

$E_{n} = - 2.18 \times 10^{- 18} \frac{Z^{2}}{n^{2}} J$

For first Bohr orbit of $H e^{+}$ ,

$E_{{He}^{+}} = - 2.18 \times 10^{- 18} \times \frac{4}{1} = - 8.72 \times 10^{- 18} J$

For first Bohr orbit of $L i^{2 +}$ ,

$E_{{Li}^{2 +}} = - 2.18 \times 10^{- 18} \times \frac{9}{1} = - 19.62 \times 10^{- 18} J$

Bohr’s radius, $r_{n} = 52.9 \times \frac{n^{2}}{Z} pm$

For first Bohr orbit of $H e^{+}$ ,

$r_{({He}^{+})} = 52.9 \times \frac{1}{2} = 26.45$ $pm$

For first Bohr orbit of $L i^{2 +}$ ,

$r_{({Li}^{2 +})} = 52.9 \times \frac{1}{3} = 17.63$ $pm$

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