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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 = $0.53\mathring{A}$).
According to Bohr's quantisation of angular momentum
$m v r=\frac{n h}{2 \pi}$
or $\frac{h}{m v}=\frac{2 \pi r}{n}$...(i)
de- Broglie wavelength
$\lambda=\frac{h}{m v}$...(ii)
From Eqs. (i) and (ii), we get
Wavelength $\lambda=\frac{2 \pi r}{n}$
$=\frac{2 \times \pi \times 0.53\,\mathring{A}}{1}$
$=3.33\, \mathring{A}$