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Topic: Bohr model and its limitations· Atomic Structure

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]

1. \(E_n(\mathrm{Li^{2+}}) = -19.62\times10^{-18}\) J; \(r_n(\mathrm{Li^{2+}}) = 17.6\) pm; \(E_n(\mathrm{He^+}) = -8.72\times10^{-18}\) J; \(r_n(\mathrm{He^+}) = 26.4\) pm

2. \(E_n(\mathrm{Li^{2+}}) = -8.72\times10^{-18}\) J; \(r_n(\mathrm{Li^{2+}}) = 26.4\) pm; \(E_n(\mathrm{He^+}) = -19.62\times10^{-18}\) J; \(r_n(\mathrm{He^+}) = 17.6\) pm

3. \(E_n(\mathrm{Li^{2+}}) = -19.62\times10^{-16}\) J; \(r_n(\mathrm{Li^{2+}}) = 17.6\) pm; \(E_n(\mathrm{He^+}) = -8.72\times10^{-16}\) J; \(r_n(\mathrm{He^+}) = 26.4\) pm

4. \(E_n(\mathrm{Li^{2+}}) = -8.72\times10^{-16}\) J; \(r_n(\mathrm{Li^{2+}}) = 17.6\) pm; \(E_n(\mathrm{He^+}) = -19.62\times10^{-16}\) J; \(r_n(\mathrm{He^+}) = 17.6\) pm

1 answer

Best answer

Answer: 1. \(E_n(\mathrm{Li^{2+}}) = -19.62\times10^{-18}\) J; \(r_n(\mathrm{Li^{2+}}) = 17.6\) pm; \(E_n(\mathrm{He^+}) = -8.72\times10^{-18}\) J; \(r_n(\mathrm{He^+}) = 26.4\) pm

Key idea: for hydrogen-like ions \(E_n=-R_H\frac{Z^2}{n^2}\) and \(r_n=a_0\frac{n^2}{Z}\).

  • \(\mathrm{Li^{2+}}\) (Z = 3): \(E=-2.18\times10^{-18}\times9=-19.62\times10^{-18}\) J; \(r=52.9/3=17.6\) pm.
  • \(\mathrm{He^+}\) (Z = 2): \(E=-2.18\times10^{-18}\times4=-8.72\times10^{-18}\) J; \(r=52.9/2=26.4\) pm.

Option (1). Others swap the ions or use wrong powers of ten.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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