Question

The electron of a hydrogen atom is in its $n$th Bohr orbit and has a de-Broglie wavelength of 13.4 $\stackrel{0}{A}$? The value of $n$ is:
[$h=6.63\times {10}^{-34}{m}^{2}kg{s}^{-1}$]

Answer 1

de-Broglie eqution, $\lambda =\frac{h}{mv}$
$v=\frac{2.18\times {10}^6}{n}\times Zm/s$
$v=\frac{2.18\times {10}^6}{n}m/s\text{Since}[Z=1]$
$\lambda =\frac{6.63\times {10}^{-34}\times n}{9.1\times {10}^{-31}\times 2.18\times {10}^6}$
$\Rightarrow 13.4\times {10}^{-10}=\frac{6.63\times {10}^{-34}\times n}{9.1\times {10}^{-31}\times 2.18\times {10}^6}$

$\Rightarrow n=\frac{9.1\times 2.18\times 13.4}{6.63}\times {10}^{-1}\approx 4$