Physics - Modern Physics Question with Solution | TestHub

According to Bohr's quantization condition

$mvr=\frac{nh}{2\pi }$

$\therefore \frac{h }{mv}= \frac{2\pi r}{n}$... (i)

∵  de-Broglie wavelength

$\lambda =\frac{h}{mv}$... (ii)

From Eqs. (i) and (ii),

$\lambda =\frac{2\pi r}{n}$

∵  Electron in the ground state

$r=\mathrm{0.53 }\stackrel{\circ }{\mathrm{A}}$and  n = 1

$\Rightarrow \lambda =\frac{2\pi \times 0.53 {\displaystyle \stackrel{\circ }{\mathrm{A}}}}{1}$
$=\mathrm{3.33 }\stackrel{\circ }{\mathrm{A}}$