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NEET Chemistry QUESTION #1095
Question 1
Energy and radius of first Bohr orbit of He\(^+\) and Li\(^{2+}\) are: [Given: \(R_H = 2.18\times10^{-18}\ \text{J}\), \(a_0=52.9\ \text{pm}\)]
  • \(E_n(\text{Li}^{2+})=-19.62\times10^{-16}\ \text{J}\); \(r_n(\text{Li}^{2+})=17.6\ \text{pm}\); \(E_n(\text{He}^+)=-8.72\times10^{-16}\ \text{J}\); \(r_n(\text{He}^+)=26.4\ \text{pm}\)
  • \(E_n(\text{Li}^{2+})=-8.72\times10^{-16}\ \text{J}\); \(r_n(\text{Li}^{2+})=17.6\ \text{pm}\); \(E_n(\text{He}^+)=-19.62\times10^{-16}\ \text{J}\); \(r_n(\text{He}^+)=17.6\ \text{pm}\)
  • \(E_n(\text{Li}^{2+})=-19.62\times10^{-18}\ \text{J}\); \(r_n(\text{Li}^{2+})=17.6\ \text{pm}\); \(E_n(\text{He}^+)=-8.72\times10^{-18}\ \text{J}\); \(r_n(\text{He}^+)=26.4\ \text{pm}\)✔️
  • \(E_n(\text{Li}^{2+})=-8.72\times10^{-18}\ \text{J}\); \(r_n(\text{Li}^{2+})=26.4\ \text{pm}\); \(E_n(\text{He}^+)=-19.62\times10^{-18}\ \text{J}\); \(r_n(\text{He}^+)=17.6\ \text{pm}\)
Correct Answer Explanation
\(E_n = -Z^2 R_H/n^2\). For Li\(^{2+}\) (\(Z=3, n=1\)): \(E=-9\times2.18\times10^{-18}=-19.62\times10^{-18}\ \text{J}\). For He\(^+\) (\(Z=2, n=1\)): \(E=-4\times2.18\times10^{-18}=-8.72\times10^{-18}\ \text{J}\). Radii: \(r=a_0 n^2/Z\). Li\(^{2+}\): \(r=52.9/3=17.6\ \text{pm}\). He\(^+\): \(r=52.9/2=26.45\approx26.4\ \text{pm}\). Correct: option (3).