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Q.
For hydrogen atom electron in $n ^{\text {th }}$ Bohr orbit, the ratio of radius of orbit to its de-Broglie wavelength is
Haryana PMTHaryana PMT 2009Dual Nature of Radiation and Matter
Solution:
For nth Bohr orbit, $r=\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m Z e^{2}}$
de-Broglie wavelength $\lambda=\frac{h}{m v}$
Ratio of both $r$ and $\lambda$,
we have $\frac{r}{\lambda}=\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m Z e^{2}} \times \frac{m v}{h}$
$=\frac{\varepsilon_{0} n^{2} h v}{\pi Z e^{2}}$
But $v=\frac{Z e^{2}}{2 h \varepsilon_{0} n}$ for nth orbit
Hence, $\frac{r}{\lambda}=\frac{n}{2 \pi}$