Question

Explain the quantization of angular momentum to behave an electron as a wave in an atom.

Answer 1

Hint: To explain the quantization of angular momentum to behave an electron as wave in a atom, we have to use the concept of Angular momentum of electron given by Bohr’s atomic model which states that the angular momentum of electron orbiting around the nucleus is quantized. We also use the concept of de Broglie Hypothesis.

Complete step by step answer:
The idea about angular momentum of an electron is given by Bohr which is $$mvr$$ or ${\displaystyle \frac{nh}{2\pi }}$ (where $v$ is the velocity, $n$ is the orbit in which electron is, $m$ is mass of the electron, and $r$ is the radius of the $n^{th}$ orbit).

According to Bohr’s atomic model, the electron in a hydrogen atom revolves in a circular orbit around the nucleus with the nucleus at the center of orbit. Also, The electron revolves around the nucleus only in those orbits for which the angular momentum is equal to an integral multiple of ${\displaystyle \frac{h}{2\pi }}$ , where $h$ is Planck’s constant.

Now, the angular momentum of an electron revolving around the nucleus is given by
$L=I\omega$
Where, $I$ is the moment of inertia of an electron about its axis of revolution and $\omega$ is angular velocity given by $\omega ={\displaystyle \frac{v}{r}}$ .
So, $L=m{r}^2\left({\displaystyle \frac{v}{r}}\right)=mvr$
And According to Bohr, angular momentum $L=n{\displaystyle \frac{h}{2\pi }}$
 $mvr=n{\displaystyle \frac{h}{2\pi }}$ , where $n=1,2,3,4,....$
That is, electrons revolve only in those orbits that satisfy the above equation.

Further, de Broglie extends the concept of angular momentum.According to de Broglie hypothesis, the wavelength of the electron is given by
$\lambda ={\displaystyle \frac{h}{p}}$
Where, $p$ - momentum of the electron given by $p=mv$
De Broglie suggested that Particle waves can lead to standing waves. When a stationary string is plucked, a number of wavelengths are excited and we know that only those wavelengths survive which form a standing wave in the string. Thus, standing waves are formed only when the total distance travelled by a wave is an integral number of wavelengths.

Hence, for any electron moving in $n^{th}$circular orbit of radius $r_n$, the total distance is equal to the circumference of the orbit, $2\pi r_n$ .
So, $2\pi r_n=n\lambda$
Substituting for $\lambda$ , we get
$2\pi r_n={\displaystyle \frac{nh}{mv}}$
$\therefore mv{r}_n={\displaystyle \frac{nh}{2\pi }}$
Hence, de Broglie hypothesis successfully proved Bohr’s postulate of the quantization of angular momentum of the orbiting electron. Also, the quantized electron orbits and energy states are due to the wave nature of the electron.

Note: Bohr’s atomic model does not explain the spectra of atoms having more than one electron and hence it is limited for the atoms like hydrogen. The value of $n$ defines the principal quantum number. We should note standing waves are formed only when the total distance travelled by a wave is an integral number of wavelengths.