Question

An electron in an excited state of $L{{i}^{2+}}$ ion has angular momentum $\dfrac{3h}{2\pi }$. The de-Broglie wavelength of the electron in this state is $p\pi {{\alpha }_{0}}$ (where ${{\alpha }_{0}}$ is the Bohr radius). The value of p is :

Answer 1

Use the formula for the angular momentum of an electron orbiting in nth orbit and find the value of n. They use the formula for angular momentum from the mechanics chapter and equate the two. Then use the formula for wavelength of the electron and substitute for the linear momentum. Finally use the formula for the radius of the orbit in which the electron is revolving and find the value of p.

Formula used:
$L=\dfrac{nh}{2\pi }$
L = mvr
$\lambda =\dfrac{h}{mv}$
$r=\dfrac{{{\alpha }_{0}}{{n}^{2}}}{Z}$

Complete answer:
The angular momentum of an electron orbiting around the nucleus in orbit n is given as $L=\dfrac{nh}{2\pi }$, where h is the Planck’s constant.
It is given that $L=\dfrac{3h}{2\pi }$.
This means that n = 3.
We know that when a particle is moving in a circular path of radius r is given as L = mvr, where m is the mass of the particle and v is its velocity.
In this case, the particle is an electron therefore, we get that
$L=\dfrac{3h}{2\pi }=mvr$
$\Rightarrow mv=\dfrac{3h}{2\pi r}$ ….. (i)
An electron also has a wave nature and the wavelength of an electron is given as $\lambda =\dfrac{h}{mv}$.......(ii).
Substitute the value of mv in (ii).
$\Rightarrow \lambda =\dfrac{h}{\dfrac{3h}{2\pi r}}$
$\Rightarrow \lambda =\dfrac{2\pi r}{3}$
But it is given that $\lambda =p\pi {{\alpha }_{0}}$.
$\Rightarrow p\pi {{\alpha }_{0}}=\dfrac{2\pi r}{3}$ …. (iii).
The radius of the orbit of the electron is given as $r=\dfrac{{{\alpha }_{0}}{{n}^{2}}}{Z}$, Z is the atomic number and ${{\alpha }_{0}}$ is the Bohr radius.
In this case, n=3 and for lithium Z = 3.
$\Rightarrow r=\dfrac{{{\alpha }_{0}}{{3}^{2}}}{3}=3{{\alpha }_{0}}$
Substitute the value of r in (iii).
$\Rightarrow p\pi {{\alpha }_{0}}=\dfrac{2\pi ({{\alpha }_{0}})}{3}$
$\Rightarrow p=2$
Therefore, the value of p is 2.

Note:
Bohr’s radius is equal to the radius of the orbit of the electron of a hydrogen atom in ground state. This is because the atomic number of hydrogen is one. Hence, Z = 1. And since the electron is in ground state, n = 1.
When we say that the electron is in ground state, it is revolving in the first orbit. When we say that the electron is in an excited state, it is revolving in an orbit other than the first orbit.