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
$L=mvr=\frac{nh}{2\pi }$
de-Broglie wavelength, $\lambda =\frac{h}{mv}$
$\Rightarrow \frac{mv}{h}=\frac{1}{\lambda }$
$\Rightarrow \frac{r}{\lambda }=\frac{n}{2\pi }$
$\Rightarrow \lambda =\frac{2\pi r}{n}$
Now, $r\propto n^2$ (Bohr radius, $r=0.529\times n^2\mathring{A}$)
$\Rightarrow \lambda \propto n$