Q. The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is (Radius of the first orbit of hydrogen atom =$0.53\mathring{\mathrm{A}}$ )
A) 1.67$\ \mathring{\mathrm{A}}$
B) 3.33$\ \mathring{\mathrm{A}}$
C) 1.06$\ \mathring{\mathrm{A}}$
D) 0.53$\ \mathring{\mathrm{A}}$
Explanation
(b) According to Bohr’s quantisation of angular momentum
$mvr = \frac{nh}{2\pi}$
Or $\ \ \ \ \ \ \ \ \ \ \frac{h}{mv} = \frac{2\pi r}{n}\ \ \ ..(i)$
de-Broglie wavelength
$\ \lambda = \frac{h}{mv}\ \ \ \ ..(ii)$ From Eqs. (i) and (ii), we get
Wavelength $\lambda = \frac{2\pi r}{n}$
=$\frac{2 \times \pi \times 0.53\ \mathring{\mathrm{A}}}{1} = 3.33\mathring{\mathrm{A}}\ \ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.