Question

Energy of separation of $e^{-}$ in hydrogen like atom in excited state is $3.4ev$. The de broglie wavelength in $\mathring{A}$ associated with $e^{-}$ is? Given rad of $1th$ orbit of $H$ atom is $0.53\mathring{A}$.

Answer 1

$\lambda =\frac{h}{p}$(de-broglie)
$P=mv$
$P^2=m^2{v}^2$
$\frac{P^2}{2m}\frac{1}{2}m{v}^2$
$=KE$
$2mKE=P^2$
$\Rightarrow P=\sqrt{2m\times KE}$
$\lambda =\frac{h}{\sqrt{2m,KE}}$