A) \[\frac{n}{2\pi }\]
B) \[\frac{{{n}^{2}}}{2\pi }\]
C) \[\frac{1}{2\pi n}\]
D) \[\frac{1}{2\pi {{n}^{2}}}\]
Correct Answer: A
Solution :
For nth Bohr orbit, \[r=\frac{{{\varepsilon }_{0}}{{n}^{2}}{{h}^{2}}}{\pi mZ{{e}^{2}}}\] de-Broglie wavelength \[\lambda =\frac{h}{mv}\] Ratio of both r and \[\lambda \], we have \[\frac{r}{\lambda }=\frac{{{\varepsilon }_{0}}{{n}^{2}}{{h}^{2}}}{\pi mZ{{e}^{2}}}\times \frac{mv}{h}\] \[=\frac{{{\varepsilon }_{0}}{{n}^{2}}hv}{\pi Z{{e}^{2}}}\] But \[v=\frac{Z{{e}^{2}}}{2h{{\varepsilon }_{0}}n}\] for nth orbit Hence, \[\frac{r}{\lambda }=\frac{n}{2\pi }\]
You need to login to perform this action.
You will be redirected in
3 sec
