Question

The de-BrogliAe wavelength $\left(\text{in}{A}^{\circ }\right)$ associated with electron in the $3^{rd}$ orbit of $H{e}^{+}$ ion is (approx)

Answer 1

$\lambda =\frac{h}{mv}$
$v=2.18\times {10}^6\times \frac{2}{3}m/s$
$\lambda =\frac{6.626\times {10}^{-34}\times 3}{2.18\times 2\times {10}^6\times 9.1\times {10}^{-31}}$
$=5.01\times {10}^{-10}m$