Search any question & find its solution
Question:
Answered & Verified by Expert
An electron in an excited state of $\mathrm{Li}^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of electron in this state is $p \pi a_{0}$ (where, $a_{0}=$ Bohr radius ). The value of $p$ is
Options:
Solution:
1686 Upvotes
Verified Answer
The correct answer is:
2
According to de-Broglie hypothesis,
$=\frac{n h}{2 \pi}=\frac{3 h}{2 \pi}=m v r$
$\Rightarrow \quad n=3$
As, wavelength, $\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h r}{m v r}$
$=\frac{h r}{3 h}=\frac{2}{3} \pi r$
For Li ${ }^{2+}$ atom, radius of orbit,
$r=r_{0} \frac{n^{2}}{Z}=a_{0} \frac{3^{2}}{3}=3 a_{0}$
$\lambda=\frac{2}{3} \pi \times a_{0} \times 3=2 \pi a_{0}=p \pi a_{0} \text { (given) }$
$\quad p=2$
$=\frac{n h}{2 \pi}=\frac{3 h}{2 \pi}=m v r$
$\Rightarrow \quad n=3$
As, wavelength, $\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h r}{m v r}$
$=\frac{h r}{3 h}=\frac{2}{3} \pi r$
For Li ${ }^{2+}$ atom, radius of orbit,
$r=r_{0} \frac{n^{2}}{Z}=a_{0} \frac{3^{2}}{3}=3 a_{0}$
$\lambda=\frac{2}{3} \pi \times a_{0} \times 3=2 \pi a_{0}=p \pi a_{0} \text { (given) }$
$\quad p=2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.