Question

Hydrogen atom is in its $n^{\text {th }}$ energy state. If de-Broglie wavelength of the electron is $\lambda$, then

• A. $\lambda \propto \frac{1}{n^{2}}$
• B. $\lambda \propto \frac{1}{n}$
• C. $\lambda \propto n^2 $
• D. $\lambda \propto n $

Answer 1

Answer: (D)

Angular momentum of electron in $n^{\text {th }}$ orbit of hydrogen is
$L=m v r=\frac{n h}{2 \pi}$
de-Broglie wavelength, $\lambda=\frac{h}{m v}$
$\Rightarrow \frac{m v}{h}=\frac{1}{\lambda}$
$\Rightarrow \frac{r}{\lambda}=\frac{n}{2 \pi}$
$\Rightarrow \lambda=\frac{2 \pi r}{n}$
Now, $r \propto n^{2}$ (Bohr radius, $r=0.529 \times n^{2}\, \mathring{A}$)
$\Rightarrow \lambda \propto n$