Question

The radius of the first Bohr orbit of hydrogen atom is $0.59\stackrel{o}{A}$. The radius of the third orbit of $H{e}^{+}$ will be:

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

Answer 1

Answer: (D)

$2.38\stackrel{o}{A}$

Let us consider the problem:

Radius of $n$th Bohr orbit, $r_n=\frac{n^2{h}^2}{4{\pi }^{2}mZ{e}^2}$

For hydrogen atom $Z=1$, first orbit $n=1$

$r_1=\frac{h^2}{4{\pi }^{2}m{e}^2}=0.529A^{\circ }$

For $H{e}^{+}$,$Z=2,thirdorbit,$n=3$

$r_3\left(H{e}^{+}\right)=\frac{3^2{h}^2}{4{\pi }^{2}m\times 2\times e^2}$

$r_3\left(H{e}^{+}\right)=\frac{3^2{h}^2}{4{\pi }^{2}m\times 2\times e^2}$

$=\frac{9}{2}\left[\frac{h^2}{4{\pi }^{2}m{e}^2}\right]=\frac{9}{2}\times 0.529=2.380A\circ$