Question

please answer quickly

Answer 1

de Broglie wavelength is inversely proportionate to angular momentum of an electron.

$\lambda \propto \dfrac{1}{L}$
as, we know,

$\lambda = \dfrac{h}{p}$
and $\vec{L} = \vec{r} \times \vec{p}$

And, according to Bohr's model angular momentum is $\vec{L} = \dfrac{nh}{2 \pi}$

So, $L \propto n$

So, as $\lambda \propto \dfrac{1}{L}$, $\lambda \propto \dfrac{1}{n}$

n=number of orbital
p=linear momentum
L=angular momentum
$\lambda$ = wavelength
r=position vector/displacement
h=Planck's Constant