Question

If the radius of the first orbit of hydrogen atom is $0.53A^{\circ }$, then the de-Broglie wavelength of the electron in the ground state of hydrogen atom will be :

• A. $0.53A^{\circ }$
• B. $3.33A^{\circ }$
• C. $1.67A^{\circ }$
• D. $1.06A^{\circ }$

Answer 1

The correct option is B $3.33A^{\circ }$

According to Bohr's quantization condition
$mvr=\frac{nh}{2\pi }$
$\therefore \frac{h}{mv}=\frac{2\pi r}{n}$ ....(i)
$\because$ de-Broglie wavelength of a particle
$\lambda =\frac{h}{mv}$ ....(ii)
From Eqs. (i) and (ii),
$\lambda =\frac{2\pi r}{n}$
$\because$ Electron in ground state
$r=0.53A^{\circ }$ and n=1
$\Rightarrow \lambda =\frac{2\pi \times 0.53A^{\circ }}{1}$$=3.33A^{\circ }$