Question

The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is ......... (radius of the first orbit of hydrogen atom = $0.53\mathring{A}$).

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

Answer 1

Answer: (D)

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 $\lambda =\frac{2\pi r}{n}$
$=\frac{2\times \pi \times 0.53\mathring{A}}{1}$
$=3.33\mathring{A}$