Question

The frequency of $e^{-}$ in the first Bohr orbit in a H-atom is:

• A. $6530\times {10}^{12}Hz$
• B. $6840\times {10}^{12}Hz$
• C. $6430\times {10}^{12}Hz$
• D. $6734\times {10}^{12}Hz$

Answer 1

Answer: (A)

$6530\times {10}^{12}Hz$

Velocity in Bohr's first orbit is given by $mvr=\frac{nh}{2\pi }$
For Bohr's first orbit $n=1$
$v=\frac{h}{2\pi mr}$
Radius $r$ is given by, $r_n=0.53\times {10}^{-10}\times n^{2}m$
For Bohr's first orbit $n=1$
$r=0.53\times {10}^{-10}m$
Frequency $\nu$ is given by $\nu =\frac{v}{2\pi r}=\frac{h}{4{\pi }^{2}m{r}^2}$
Mass of electron $m=9.1096\times {10}^{-31}kg$
$\Rightarrow \nu =\frac{6.62\times {10}^{-34}}{4{\pi }^2\times 9.1096\times {10}^{-31}\times {(0.53\times {10}^{-10})}^2}=6530\times {10}^{12}Hz$