An expression for the frequency of revolution of the electron in the Bohr's orbit is :
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : $\frac{me^4}{4\epsilon^2_0h^3}\times \frac{1}{n^3}$
In Bohr's orbit
$mvr=\frac{nh}{2\pi}$ ............. (i)
$\frac{e^2z}{4\pi \epsilon r^2}=\frac{mv^2}{r}$
$mv^2r=\frac{e^2z}{4\pi \epsilon_0}$ ............ (ii)
On putting the value of mvr from equation (i)
$v\times \frac{nh}{2\pi }\times \frac{e^2z}{4\pi \epsilon_0}$
$v=\left(\frac{e^2z}{2nh}\right)$
$\frac{m\times e^2z}{2nh}z=\frac{nh}{2\pi}$
$r=\frac{n^2h^2}{\pi me^2z}$
Now, $f=\frac{v}{2\pi r}$
$=\frac{e^2z}{\frac{2nh}{\frac{2\pi n^2h^2}{\pi me^2z}}}$
$=\frac{me^4}{4\epsilon^2_0h^3}\times \frac{1}{n^3}$