Question

The circumference of the first Bohr orbit in H atom is $3.322\times {10}^{-10}m$. What is the velocity of the electron of this orbit?

• A. 2.19 × 10⁶ m/s
• B. 4.19 × 10⁶ m/s
• C. 3.57 × 10⁶ m/s
• D. 6.00 × 10⁶ m/s

Answer 1

The correct option is A

2.19 × 10⁶ m/s

From Bohr's postulates, we know that

$\to$ An electron can move only in those orbits for which its angular momentum is an integral multiple of $\frac{h}{2\pi }$.

$\Rightarrow$ $m_{e}vr$ = $\frac{nh}{2\pi }$

Now,

According to Bohr's model,

$mvr=\frac{nh}{2\pi }$;

Given that $n$ = $1$ & $2\pi r$ (circumference)

= $3.322\times {10}^{-10}m$

$v=\frac{nh}{2\pi mr}=\frac{1\times (6.626\times {10}^{-34}Js)}{(9.1\times {10}^{-31}kg)\times 3.322\times {10}^{-10}m}$

= $2.19\times {10}^{6}m{s}^{-1}$