Nuclear Motion Slide

PHY-1109 Physics

Nuclear Motion and Reduced mass

The nuclear mass affects the wavelengths of spectral lines

(There is a figure, Fig. 1, illustrating the electron and nucleus revolving around a common center of mass) Fig. 1: Both the electron and nucleus of a hydrogen atom revolve around a common center of mass.

In developing the theory of hydrogen atom, its nucleus remains stationary while the orbital electron revolves around it. This is not an unreasonable assumption: the proton mass is 1836 times greater than the electron mass and the electrostatic force $F_e=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}\text{(1)}$

We begin in expressing in a different way the condition that a stable orbit in an atom consist of an integral number of electron de Broglie wavelengths.

This condition states that $n\lambda =2\pi r\text{(2)}$

Since the de Broglie wavelength $\lambda$ is given by $\lambda =\frac{h}{mv}\text{(3)}$

We may equivalently write $mvr=\frac{nh}{2\pi }=n\hslash \text{(4)}$

In terms of the angular velocity of the electron, the quantization rule is stated $m\omega r^2=n\hslash \text{(5)}$

Since $\omega =\frac{v}{r}\text{(6)}$

If the nucleus has a mass $M$ that is not infinite, both it and its orbital electron revolve around a common center of mass. If we think of the nucleus and electron in a hydrogen atom as being at opposite ends of a massless rod $r$ long, in effect constituting a lopsided dumbbell, the center of mass, which is $r_N$ distant from the nucleus and $r_E$ distant from the electron, may be found from the requirement $m{r}_E=M{r}_N\text{(7)}$

Where $r=r_E+r_N\text{(8)}$

The total angular momentum of the hydrogen atom is the sum of the angular momentum of the electron and that of the nucleus $\text{Total angular momentum}=\text{electron angular momentum}+\text{nuclear angular momentum}$ $=m\omega r_E^2+M\omega r_N^2$

The angular velocity $\omega$ is the same for both particles.

According to Bohr’s first postulate, the total angular momentum of the atom must be an integral multiple of $\hslash$, and so $m\omega r_E^2+M\omega r_N^2=n\hslash \text{(9)}$

From eqns. (7) & (8) we find that $r_E=\left(\frac{M}{M+m}\right)r\text{(10)}$ $r_N=\left(\frac{m}{M+m}\right)r\text{(11)}$

And so eq. (9) becomes $\left(\frac{mM}{M+m}\right)\omega r^2=n\hslash \text{(12)}$

The condition for force balance must also be generalized to take into account nuclear motion. Its proper expression is $m{\omega }^2{r}_E=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}\text{(13)}$

Substituting $r_E$ from eq. (10) we find $\left(\frac{mM}{M+m}\right){\omega }^{2}r=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}\text{(14)}$

If we let $m^{\prime }=\left(\frac{mM}{M+m}\right)\text{(15)}$

We see that eqns. (12) & (14) become respectively $m^{\prime }\omega r^2=n\hslash \text{(16)}$ $m^{\prime }{\omega }^{2}r=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}\text{(17)}$

From eqns. (16) & (17) we can obtain an expression for the energy levels of the hydrogen atom.

The result is $E_n=-\frac{m^{\prime }{e}^4}{8{ϵ}_o^2{h}^2}\left(\frac{1}{n^2}\right)\text{(18)}$

Here $m^{\prime }$ replaces the electron mass $m$. The quantity $m^{\prime }$ is known as reduced mass. Due to the motion of the nucleus, all the energy levels of the hydrogen are changed by the fraction $\frac{m^{\prime }}{m}=\left(\frac{M}{M+m}\right)$ $=\frac{1836}{1837}=0.99945$

An increase of $0.055$ percent since the energies $E_n$, being smaller in absolute value, are therefore less negative.

The value of the Rydberg number, $R$ to eight significant figures without correcting for nuclear motion is $1.0973731\times 10^7{\text{ m}}^{-1}$; the correction lowers it to $1.0967758\times 10^7{\text{ m}}^{-1}$.

Because of the greater nuclear mass, the spectral lines of deuterium are all shifted slightly to wavelengths shorter than the corresponding ones of ordinary hydrogen.

Thus the $H_{\alpha }$ line of deuterium, which arises at a transition from the $n=3$ to the $n=2$ energy level, occurs at a wavelength of $6561\text{ A˚}$, whereas the $H_{\alpha }$ line of hydrogen occurs at $6563\text{ A˚}$. This difference in wavelength was responsible for the identification of deuterium in 1932 by the American chemist Harold Urey.

Hydrogenic Atoms

The nucleus of a hydrogenic atom of atomic number $Z$ carries the charge $+Ze$, and so the electrostatic force it exerts on the orbital electron is $F_e=\frac{1}{4\pi {ϵ}_o}\frac{Z{e}^2}{r^2}\text{(19)}$

Consequently the formula for the energy levels of a hydrogenic atom is $E_n=-\frac{m^{\prime }{Z}^2{e}^4}{8{ϵ}_o^2{h}^2}\left(\frac{1}{n^2}\right)\text{(20)}$

The general formula for the spectral lines of a hydrogenic atom of atomic number $Z$ is $\frac{1}{\lambda }=\frac{m^{\prime }{Z}^2{e}^4}{8{ϵ}_o^{2}c{h}^3}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\text{(21)}$

(There is a figure, Fig. 2, showing energy levels for Hydrogen (H) and singly ionized helium (${\text{He}}^{+}$)) Fig. 2: Energy levels in hydrogen and singly ionized helium.