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

Answer: 2

$2\pi r=n\lambda$
$\therefore \lambda =\frac{2\pi r}{n}=\frac{2\times 3.14\times a_0{n}^2}{n\times Z}$
$\therefore \lambda =\frac{2\times 3.14\times 0.529\times n}{Z}$
$\therefore n=\frac{\lambda \times Z}{2\times 3.14\times 0.529}$
$=\frac{6.65\times 1}{2\times 3.14\times 0.529}=2$