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 K, we have \[\frac{r}{\lambda }=\frac{{{\varepsilon }_{0}}{{n}^{2}}{{h}^{2}}}{\pi nZ{{e}^{2}}}\times \frac{mv}{h}\] \[=\frac{{{\varepsilon }_{0}}{{n}^{2}}hv}{\pi Z{{e}^{2}}}\] But \[v=\frac{Z{{e}^{2}}}{2h{{\varepsilon }_{0}}n}\]for nth orbit Hence, \[\frac{r}{\lambda }=\frac{n}{2\pi }\]
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