Question

The velocity of an electron in the first Bohr orbit of a hydrogen atom is equal to the velocity of the electron in the $n^{th}$ Bohr orbit of $B{e}^{3+}$ atom. Find the value of n.

• A. 8
• B. 4
• C. 16
• D. 2

Answer 1

The correct option is B 4

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-atom,
Z = 4 and n is unknown for $B{e}^{3+}$.
Equating the values, we get
$\therefore \frac{1}{1}=\frac{4}{n}$
$n=4$