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ChemistryNEET 2025Bohr model and hydrogen spectrum

Structure of Atom

The energy and radius of the first Bohr orbit of He+\mathrm{He^+} and Li2+\mathrm{Li^{2+}} are (given RH=2.18×10−18R_H = 2.18 \times 10^{-18} J, a0=52.9a_0 = 52.9 pm):

  1. A.E(Li2+)=−19.62×10−18E(\mathrm{Li^{2+}}) = -19.62 \times 10^{-18} J, r(Li2+)=17.6r(\mathrm{Li^{2+}}) = 17.6 pm; E(He+)=−8.72×10−18E(\mathrm{He^+}) = -8.72 \times 10^{-18} J, r(He+)=26.4r(\mathrm{He^+}) = 26.4 pm✓
  2. B.E(Li2+)=−8.72×10−18E(\mathrm{Li^{2+}}) = -8.72 \times 10^{-18} J, r(Li2+)=26.4r(\mathrm{Li^{2+}}) = 26.4 pm; E(He+)=−19.62×10−18E(\mathrm{He^+}) = -19.62 \times 10^{-18} J, r(He+)=17.6r(\mathrm{He^+}) = 17.6 pm
  3. C.E(Li2+)=−19.62×10−16E(\mathrm{Li^{2+}}) = -19.62 \times 10^{-16} J, r(Li2+)=17.6r(\mathrm{Li^{2+}}) = 17.6 pm; E(He+)=−8.72×10−16E(\mathrm{He^+}) = -8.72 \times 10^{-16} J, r(He+)=26.4r(\mathrm{He^+}) = 26.4 pm
  4. D.E(Li2+)=−8.72×10−16E(\mathrm{Li^{2+}}) = -8.72 \times 10^{-16} J, r(Li2+)=17.6r(\mathrm{Li^{2+}}) = 17.6 pm; E(He+)=−19.62×10−16E(\mathrm{He^+}) = -19.62 \times 10^{-16} J, r(He+)=17.6r(\mathrm{He^+}) = 17.6 pm

Solution

En=−RHZ2n2E_n = -R_H\frac{Z^2}{n^2} and rn=a0n2Zr_n = a_0\frac{n^2}{Z}. He+\mathrm{He^+} (Z=2Z=2): E=−4×2.18×10−18=−8.72×10−18E = -4 \times 2.18\times10^{-18} = -8.72\times10^{-18} J, r=52.9/2=26.4r = 52.9/2 = 26.4 pm. Li2+\mathrm{Li^{2+}} (Z=3Z=3): E=−9×2.18×10−18=−19.62×10−18E = -9 \times 2.18\times10^{-18} = -19.62\times10^{-18} J, r=52.9/3=17.6r = 52.9/3 = 17.6 pm.

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