Question

The de Broglie wavelength of an electron in a certain Bohr's orbit of Hydrogen atom is $6.65\,\mathring{A}$. The principal quantum number of orbit is______.
[Given: $\pi=3.14$ and $a _{0}=0.529\,\mathring{A}$ ]

Answer 1

$2 \pi r = n \lambda$
$\therefore \lambda =\frac{2 \pi r }{ n }=\frac{2 \times 3.14 \times a _{0} n ^{2}}{ n \times Z }$
$\therefore \lambda =\frac{2 \times 3.14 \times 0.529 \times n }{ Z }$
$\therefore n =\frac{\lambda \times Z }{2 \times 3.14 \times 0.529}$
$=\frac{6.65 \times 1}{2 \times 3.14 \times 0.529}=2$