Question

For hydrogen atom electron in $n^{\text{th }}$ Bohr orbit, the ratio of radius of orbit to its de-Broglie wavelength is

• A. $\frac{n}{2\pi }$
• B. $\frac{n^2}{2\pi }$
• C. $\frac{1}{2\pi n}$
• D. $\frac{1}{2\pi n^2}$

Answer 1

Answer: (A)

For nth Bohr orbit, $r=\frac{{\epsilon }_0{n}^2{h}^2}{\pi mZ{e}^2}$
de-Broglie wavelength $\lambda =\frac{h}{mv}$
Ratio of both $r$ and $\lambda$,
we have $\frac{r}{\lambda }=\frac{{\epsilon }_0{n}^2{h}^2}{\pi mZ{e}^2}\times \frac{mv}{h}$
$=\frac{{\epsilon }_0{n}^{2}hv}{\pi Z{e}^2}$
But $v=\frac{Z{e}^2}{2h{\epsilon }_{0}n}$ for nth orbit
Hence, $\frac{r}{\lambda }=\frac{n}{2\pi }$