The energy and radius of the first Bohr orbit of \mathrm{He^+} and \mathrm{Li^{2+}} are (given RH = 2.18 \times 10^{-18} J, a0 = 52.9 pm):

A. En = -RH\frac{Z^2}{n^2} and rn = a0\frac{n^2}{Z}. \mathrm{He^+} (Z=2): E = -4 \times 2.18\times10^{-18} = -8.72\times10^{-18} J, r = 52.9/2 = 26.4 pm. \mathrm{Li^{2+}} (Z=3): E = -9 \times 2.18\times10^{-18} = -19.62\times10^{-18} J, r = 52.9/3 = 17.6 pm.

Structure of Atom — NEET 2025 Chemistry question, solved

Structure of Atom

The energy and radius of the first Bohr orbit of $H e^{+}$ and $L i^{2 +}$ are (given $R_{H} = 2.18 \times 1 0^{- 18}$ J, $a_{0} = 52.9$ pm):

A. $E (L i^{2 +}) = - 19.62 \times 1 0^{- 18}$ J, $r (L i^{2 +}) = 17.6$ pm; $E (H e^{+}) = - 8.72 \times 1 0^{- 18}$ J, $r (H e^{+}) = 26.4$ pm✓
B. $E (L i^{2 +}) = - 8.72 \times 1 0^{- 18}$ J, $r (L i^{2 +}) = 26.4$ pm; $E (H e^{+}) = - 19.62 \times 1 0^{- 18}$ J, $r (H e^{+}) = 17.6$ pm
C. $E (L i^{2 +}) = - 19.62 \times 1 0^{- 16}$ J, $r (L i^{2 +}) = 17.6$ pm; $E (H e^{+}) = - 8.72 \times 1 0^{- 16}$ J, $r (H e^{+}) = 26.4$ pm
D. $E (L i^{2 +}) = - 8.72 \times 1 0^{- 16}$ J, $r (L i^{2 +}) = 17.6$ pm; $E (H e^{+}) = - 19.62 \times 1 0^{- 16}$ J, $r (H e^{+}) = 17.6$ pm

Solution

$E_{n} = - R_{H} \frac{Z ^{2}}{n ^{2}}$ and $r_{n} = a_{0} \frac{n ^{2}}{Z}$ . $H e^{+}$ ( $Z = 2$ ): $E = - 4 \times 2.18 \times 1 0^{- 18} = - 8.72 \times 1 0^{- 18}$ J, $r = 52.9/2 = 26.4$ pm. $L i^{2 +}$ ( $Z = 3$ ): $E = - 9 \times 2.18 \times 1 0^{- 18} = - 19.62 \times 1 0^{- 18}$ J, $r = 52.9/3 = 17.6$ pm.

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Structure of Atom

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