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NEET UGChemistryModerateNEET 2025

Energy and radius of first Bohr orbit of He+^{+} and Li2+^{2+} are
[Given RH=2.18×10−18R_H = 2.18 \times 10^{-18} J, a0=52.9_0 = 52.9 pm]

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Sol. En=−2.18×10−18×z2n2 J; rn=52.9×n2z pmE_n = \frac{-2.18 \times 10^{-18} \times z^2}{n^2} \text{ J; } r_n = \frac{52.9 \times n^2}{z} \text{ pm}

For He+^{+}

EHe+=−2.18×10−18×4=−8.72×10−18 JE_{\text{He}^+} = -2.18 \times 10^{-18} \times 4 = -8.72 \times 10^{-18} \text{ J}

rHe+=52.9×12=26.45 pmr_{\text{He}^+} = \frac{52.9 \times 1}{2} = 26.45 \text{ pm}

For Li2+^{2+}

ELi2+=−2.18×10−18×9=19.62×10−18 JE_{\text{Li}^{2+}} = -2.18 \times 10^{-18} \times 9 = 19.62 \times 10^{-18} \text{ J}

rLi2+=52.9×13=17.63 pmr_{\text{Li}^{2+}} = \frac{52.9 \times 1}{3} = 17.63 \text{ pm}

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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