Energy and radius of the first Bohr orbit of He^+ and Li^2+ are [Given RH=2.18×10^-18 J, a0=52.9 pm]:

Question

Energy and radius of the first Bohr orbit of He^+ and Li^2+ are [Given RH=2.18×10^-18 J, a0=52.9 pm]:

1. A. En(Li^2+)=-19.62×10^-18 J, rn(Li^2+)=17.6 pm; En(He^+)=-8.72×10^-18 J, rn(He^+)=26.4 pm (Correct answer)
2. B. En(Li^2+)=-8.72×10^-18 J, rn(Li^2+)=26.4 pm; En(He^+)=-19.62×10^-18 J, rn(He^+)=17.6 pm
3. C. En(Li^2+)=-19.62×10^-16 J, rn(Li^2+)=17.6 pm; En(He^+)=-8.72×10^-16 J, rn(He^+)=26.4 pm
4. D. En(Li^2+)=-8.72×10^-16 J, rn(Li^2+)=17.6 pm; En(He^+)=-19.62×10^-16 J, rn(He^+)=17.6 pm

Correct Answer

Option A — En(Li^2+)=-19.62×10^-18 J, rn(Li^2+)=17.6 pm; En(He^+)=-8.72×10^-18 J, rn(He^+)=26.4 pm

Detailed Solution & Explanation

The correct answer is Option 1.

Key Points

• $E_n=\dfrac{-2.18\times10^{-18}\,z^2}{n^2}$ J and $r_n=\dfrac{52.9\,n^2}{z}$ pm (for $n=1$).
• He$^+$ ($z=2$): $E=-2.18\times10^{-18}\times4=-8.72\times10^{-18}$ J; $r=\dfrac{52.9}{2}=26.4$ pm.
• Li$^{2+}$ ($z=3$): $E=-2.18\times10^{-18}\times9=-19.62\times10^{-18}$ J; $r=\dfrac{52.9}{3}=17.6$ pm.