. Hydrogen atom is in its $n^{\text {th }}$ energy state. If de-Broglie wavelength of the electron is…
. Hydrogen atom is in its $n^{\text {th }}$ energy state. If de-Broglie wavelength of the electron is $\lambda$, then
- $\lambda \propto \frac{1}{n^2}$
- $\lambda \propto \frac{1}{n}$
- $\lambda \propto n^2$
- $\lambda \propto n$
Solution
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 \quad \frac{r}{\lambda}=\frac{n}{2 \pi} \Rightarrow \lambda=\frac{2 \pi r}{n}
$
Now, $r \propto n^2$ (Bohr radius, $r=0.529 \times n^2 Å$ )
$
\Rightarrow \quad \lambda \propto n
$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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