Question

The radius of first Bohr orbit of hydrogen atom is $0.529\mathring{A}$. Calculate the radii of (i) the third orbit of $H{e}^{+}$ion and (ii) the second orbit of $L{i}^{2+}$ ion.

• A. $8.46\stackrel{\text{o}}{A}$
• B. $0.705\stackrel{\text{o}}{A}$
• C. $1.59\stackrel{\text{o}}{A}$
• D. $4.29\stackrel{\text{o}}{A}$
• E. $2.38\stackrel{\text{o}}{A}$

Answer 1

Answer: (B)

Let us consider the $n^{\text{th }}$ bohr orbit,
$r_n=\frac{n^2{h}^2}{4{\pi }^{2}mZ{e}^2}$
For hydrogen atom $z=1$, first orbit $n=1$
$r_1=\frac{h^2}{4\pi r^{2}m{e}^2}=0.592A^0$
(i)For $H{e}^{+}$ion, $Z=2$, third orbit, $n=3$
$r_3\left(H{e}^{+}\right)=\frac{3^2{h}^2}{4{\pi }^{2}m\times 2\times e^2}$
$=\frac{9}{2}\left[\frac{h^2}{4{\pi }^{2}m{e}^2}\right]$
$=\frac{9}{2}\times 0.592$
$=2.380A^0$
(ii) For $L{i}^{2+}$ ion, $Z=3$, second orbit $n=2$
$\begin{array}{l}{r}_2\left(L{i}^{2+}\right)=\frac{2^2{h}^2}{4{\pi }^{2}m\times 3\times e^2}\\ =\frac{4}{3}\left[\frac{h^2}{4{\pi }^{2}m{e}^2}\right]\\ =0.705A^0\end{array}$