Question

The ratio of the speed of an electron in the first orbit of hydrogen atom to that in the first orbit of $H{e}^{+}$ is

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is A $1:2$

$\begin{array}{c}\text{From single electron model we get -}\\ \begin{array}{cc}\frac{m{v}^2}{r}& =\frac{kz{e}^2}{r^2}\\ \Rightarrow m{v}^{2}r=kz{e}^2& \end{array}\\ \text{Rearranging}\to v\cdot \left(mvr\right)=kz{e}^2\end{array}$

$\begin{array}{c}\text{By Bohr quantization we have}[\text{mvr}=\frac{nh}{2\pi }]\\ \text{So putting in above eq we get,}\\ \begin{array}{c}v\cdot \frac{nh}{2\pi }=kz{e}^2\\ \Rightarrow v=\frac{2\pi kz{e}^2}{nh}\end{array}\end{array}$

$\begin{array}{c}\Rightarrow \frac{V_H}{V_{He}}=\frac{Z_H}{n_H}\times \frac{n_{He}}{Z_{He}}=\frac{1}{1}\times \frac{1}{2}=\frac{1}{2}\\ \Rightarrow \text{Velocity of hydrogen to velocity of Helium}\\ V_H:V_{He}=1:2\end{array}$