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NEET · Chemistry · STD 11 - 2. structure of atom

Energy and radius of first Bohr orbit of \(He ^{+}\)and \(Li ^{2+}\) are
\(\left[\text {Given } R_{H}=2.18 \times 10^{-18} J, a_0=52.9\text{pm}\right]\)

  1. A \(\begin{array}{l} E _{ n }\left( Li ^{2+}\right)=-19.62 \times 10^{-18} J \\ r _{ n }\left( Li ^{2+}\right)=17.6 pm \\ E _{ n }\left( He ^{+}\right)=-8.72 \times 10^{-18} J \\ r _{ n }\left( He ^{+}\right)=26.4 pm \end{array}\)
  2. B \(\begin{array}{l} E _{ n }\left( Li ^{2+}\right)=-8.72 \times 10^{-18} J \\ r _{ n }\left( Li ^{2+}\right)=26.4 pm \\ E _{ n }\left( He ^{+}\right)=-19.62 \times 10^{-18} J \\ r _{ n }\left( He ^{+}\right)=17.6 pm \end{array}\)
  3. C \(\begin{array}{l} E _{ n }\left( Li ^{2+}\right)=-19.62 \times 10^{-16} J \\ r _{ n }\left( Li ^{2+}\right)=17.6 pm \\ E _{ n }\left( He ^{+}\right)=-8.72 \times 10^{-16} J \\ r _{ n }\left( He ^{+}\right)=26.4 pm \end{array}\)
  4. D \(\begin{array}{l} E _{ n }\left( Li ^{2+}\right)=-8.72 \times 10^{-16} J \\ r _{ n }\left( Li ^{2+}\right)=17.6 pm \\ E _{ n }\left( He ^{+}\right)=-19.62 \times 10^{-16} J \\ r _{ n }\left( He ^{+}\right)=17.6 pm \end{array}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\begin{array}{l} E _{ n }\left( Li ^{2+}\right)=-19.62 \times 10^{-18} J \\ r _{ n }\left( Li ^{2+}\right)=17.6 pm \\ E _{ n }\left( He ^{+}\right)=-8.72 \times 10^{-18} J \\ r _{ n }\left( He ^{+}\right)=26.4 pm \end{array}\)

Step-by-step Solution

Detailed explanation

\(E_n=-2.18 \times 10^{-18} \times \frac{Z^2}{n^2} J \)
\( \text {For } He ^{+} \Rightarrow E _1=-2.18 \times 10^{-18} \times \frac{2^2}{1^2} J \)
\( =-8.72 \times 10^{-18} J \)
\( \text { For } Li ^{2+} \Rightarrow E _1=-2.18 \times 10^{-18} \times \frac{3^2}{1^2} J \)
\( =-19.62 \times 10^{-18} J \)
\( r _{ n }=52.9 \times \frac{ n ^2}{ Z } pm \)
\( \text {For } He ^{+} \Rightarrow r _1=52.9 \times \frac{1^2}{2} pm =26.4 pm \)
\( \text {For } Li ^{2+} \Rightarrow r _1=52.9 \times \frac{1^2}{3} pm =17.6 pm\)