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. $2.19\times {10}^{6}m/s$
• B. $4.19\times {10}^{6}m/s$
• C. $3.57\times {10}^{6}m/s$
• D. $6.00\times {10}^{6}m/s$

Answer 1

The correct option is A $2.19\times {10}^{6}m/s$

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)

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

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

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