Question

The velocity of an electron in the first Bohr orbit of a hydrogen atom is equal to the velocity of the elctron in the $n^{th}$ Bohr orbit of $H{e}^{+}$ atom. Calculate the value of $n$?

• A. $1$
• B. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is B $2$

The velocity of an electron in the $n^{th}$ Bohr orbit of a hydrogen like atom is given by:
$v_n=2.189\times {10}^6\frac{Z}{n}m/s$
$Z=1$ and $n=1$ for $H$,
$Z=2$ and $n$ is unknown for $H{e}^{+}$
Equating the values, we get
$\therefore \frac{1}{1}=\frac{2}{n}$
$\Rightarrow n=2$