Effect Of Nucleus Motion On Energy

Effect of Nucleus Motion on Energy

Till now in Bohr's model, we have assumed that all the mass of the atom is situated at the center of the atom. As the mass of the electron is very small and negligible as compared to the mass of the nucleus all the mass is assumed to be concentrated at the centre of the nucleus. But actually, the centre of mass of the nucleus-electron system is close to the nucleus as it is heavy, and to keep the centre of mass at rest, both electrons and the nucleus revolve around their centre of mass like a double star system as shown in the figure. If r is the distance of the electron from the nucleus, the distances of the nucleus and electron from the centre of mass, $r_{1}$ and $r_{2}$, can be given as

$r_1=\frac{m_{e}r}{m_N+m_e}$
and $r_2=\frac{m_{N}r}{m_N+m_e}$

We can see that the atom, nucleus, and electron revolve around their centre of mass in concentric circles of radii $r_1$ and $r_2$ to keep the centre of mass at rest. In the above system, we can analyze the motion of electrons with respect to the nucleus by assuming the nucleus to be at rest and the mass of the electron replaced by its reduced mass ${\mu }_{\mathrm{e}}$, given as -

${\mu }_{\mathrm{e}}=\frac{m_N{m}_e}{m_N+m_e}$

Now we can change our assumption and the system will look like as shown in the figure with reduced mass

Now we can derive the equation obtained by Bohr with the reduced mass also

$r_n=\frac{n^2{h}^2}{4{\pi }^{2}kZ{e}^2{m}_e}$

Now after replacing the electron mass with its reduced mass, the equation becomes

$\begin{array}{rl}{r}_n^{\prime }& =\frac{n^2{h}^2}{4{\pi }^{2}kZ{e}^2{\mu }_e}\Rightarrow r_n^{\prime }=\frac{n^2{h}^2(m_N+m_e)}{4{\pi }^{2}kZ{e}^2{m}_e{m}_N}\\ \text{or}{r}_n^{\prime }& =r_n\times \frac{m_e}{{\mu }_e}\Rightarrow r=(0.529\mathrm{A})\frac{m^2}{\mu Z}\end{array}$

But there will be no effect on the velocity because the term of mass is not present there -

$v_n=\frac{2\pi kZ{e}^2}{nh}$

Similarly for energy, we can write that

$E_n=-\frac{2{\pi }^2{k}^2{Z}^2{e}^4{m}_e}{n^2{h}^2}$

After putting the reduced mass in the equation

$\begin{array}{rl}& E_n^{\prime }=-\frac{2{\pi }^2{k}^2{Z}^2{e}^4{m}_N{m}_e}{n^2{h}^2(m_N+m_e)}\\ & E_n^{\prime }=E_n\times \frac{{\mu }_{\epsilon }}{m_e}\Rightarrow E_n=-(13.6\mathrm{eV})\frac{Z^2}{n^2}\left(\frac{\mu }{m}\right)\end{array}$

Thus, we can say that the energy of electrons will be slightly less compared to what we have derived earlier. But for numerical
calculations this small change can be neglected unless in a given problem it is asked to consider the effect of the motion of the nucleus.

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Solved Examples Based on the Effect of Nucleus Motion on Energy

Example 1: A positronium atom is a system that consists of a positron and an electron that orbit each other. The ratio of wavelength of the spectral line of positronium to that of ordinary hydrogen is:

1) 2:1

2) 4:1

3) 1:2

4) 1:4

Solution:

Effect of nucleus motion on the energy of the atom

$E_n^1=E_n\left(\frac{m_N}{m_e+m_N}\right)$
wherein
$E_n^1=$ The energy of $n^{\text{th}}$ orbital with nucleus motion
$E_n=$ The energy of ${\mathrm{n}}^{\text{th}}$ orbital without nucleus motion
$m_e=$ mass of electron
$m_N=$ mass of nucleus
since the two-particle has the same mass
$\mu =\frac{m\ast m}{m+m}=\frac{m}{2}$
since $E_n\propto m$
$\frac{E_n^{\prime }}{E_n}=\frac{1}{2}$

Hence energy level of the positronium atom is half of the corresponding energy level in the H-atom.

As a result, the wavelength in the positronium atom spectral lines is twice those of corresponding lines in the hydrogen spectrum.

Hence, the answer is the option (3).

Example 2: The wavelengths involved in the spectrum of deuterium $\left({}_1^{2}D\right)$ are slightly different from that of the hydrogen spectrum, because

1) The sizes of the two nuclei are different

2) nuclear forces are different in the two cases

3) masses of the two nuclei are different

4) The attraction between the electron and the nucleus is different in the two cases.

Solution

The wavelength of spectrum is given by $\frac{1}{\lambda }=R{z}^2(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{n_2^2})$ where $R=\frac{1.097\times {10}^7}{1+\frac{1}{nd}}$ where $m=$ mass of electron $\mathrm{M}=$ mass of nucleus

Masses of ${}_1{H}^1$ and $1{}^2$ are different. Hence the corresponding wavelengths are different.

Hence, the answer is the option (3).

Example 3: A muon is an unstable elementary particle whose mass is $207{\mathrm{m}}_{\mathrm{e}}$ and whose charge is either +e or -e. A negative muon $\left({\mu }^{\prime }\right)$ can be captured by a hydrogen nucleus (or proton) to form a muonic atom. Find the radius (in meters) of the first Bohr orbit of this atom -
1) $5\times {10}^{-13}$
2) $6.85\times {10}^{-13}$
3) $2.85\times {10}^{-13}$
4) $5.7\times {10}^{-13}$

Solution:

Here $m=207m_e$ and $M=1836m_e$; so the reduced mass is
$\mu =\frac{mM}{m+M}=\frac{\left(207m_e\right)\left(1836m_e\right)}{207m_e+1836m_e}=186m_e$

According to equation $\lambda =h/p$, the orbit radius corresponding to $n=1$ is $r_1=\frac{h^2{\epsilon }_0}{\pi m_{\epsilon }{e}^2}=5.29\times {10}^{-11}\mathrm{m}$

Hence, the radius $r^{\prime }$ that corresponds to the reduced mass $\mu$ is
$r_1^{\prime }=\left(\frac{m}{\mu }\right)r_1=\left(\frac{m_{\epsilon }}{186m_{\mathrm{e}}}\right)r_1=2.85\times {10}^{-13}\mathrm{m}$

Hence, the answer is the option (3).

Example 4: A nucleus of mass $M$ emits ${\gamma }_{\text{-ray}}$ photon of frequency ${}^{\prime }{\nu }^{\prime }$, The loss of internal energy by the nucleus is : [Take ' $c$ ' as the speed of electromagnetic wave]

1) $hv$
2) 0
3) $hv[1-\frac{hv}{2M{c}^2}]$
4) $hv[1+\frac{hv}{2M{c}^2}]$

Solution:

When the nucleus emits $\gamma$- ray there will be recoiling of the nucleus.

$p_2\to$ momentum of $\gamma -$ ray(photon)
$p_1\to$ Momentum of Nucleus
By momentum conservation,
$\begin{array}{rl}& 0=p_1+p_2\\ & 0=Mv+\frac{h}{\lambda }\\ & v=\frac{-h}{\lambda M}=\frac{-hv}{Mc}\end{array}$

Here negative sign implies the opposite direction motion of the nucleus (recoiling) when $\gamma$-ray emission takes place.

Internal energy lost by the nucleus is given to the nucleus and x-ray.

$\begin{array}{rl}\therefore \text{Internal energy lost}& ={\mathrm{KE}}_{\text{Nucleus}}+{\mathrm{E}}_{\gamma \text{-ray}}\\ & =\frac{1}{2}M{v}^2+hv\\ & =\frac{1}{2}M\frac{h^2{v}^2}{M^2{c}^2}+hv\\ & =hv[1+\left(\frac{h^2}{2M{c}^2}\right)]\end{array}$

Hence, the correct option is (4).

Example 5: A diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by ( $n$ is an integer)
$(\text{use}\hslash =\frac{h}{2\pi })$

1) $\frac{{(m_1+m_2)}^2{n}^2{\hslash }^2}{2m_1^2{m}_2^2{r}^2}$
2) $\frac{n^2{\hslash }^2}{2(m_1+m_2)r^2}$
3) $\frac{2n^2{\hslash }^2}{(m_1+m_2)r^2}$
4) $\frac{(m_1+m_2)n^2{\hslash }^2}{2m_1{m}_2{r}^2}$

Solution:

A diatomic molecule consists of two atoms of masses $m_{1}$ and $m_{2}$ at a distance $r$ apart. Let $r_{1}$and $r_{2}$ be the distances of the atoms from the centre of mass

The moment of inertia of this molecule about an axis passing through its centre of mass and perpendicular to a line joining the atoms is

The solution is correct. So no need to change it

$\begin{array}{rl}& I=m_1{r}_1^2+m_2{r}_2^2\\ & \text{As}{m}_1{r}_1=m_2{r}_2\text{or}{r}_1=\frac{m_2}{m_1}{r}_2\\ & \because r_1+r_2=r\\ & \therefore r_1=\frac{m_2}{m_1}(r-r_1)\end{array}$

On rearranging, we get

$r_1=\frac{m_{2}r}{m_1+m_2}$

similarly $r_2=\frac{m_{1}r}{m_1+m_2}$
Therefore, the moment of inertia can be written as
$\begin{array}{rl}I& =m_1{\left(\frac{m_{2}r}{m_1+m_2}\right)}^2+m_2{\left(\frac{m_{1}r}{m_1+m_2}\right)}^2\\ & =\frac{m_1{m}_2}{m_1+m_2}{r}^2\dots \cdots \cdots \cdots \cdot (i)\end{array}$

According to Bohr’s quantisation condition

$L=\frac{nh}{2\pi }$
or $L^2=\frac{n^2{h}^2}{4{\pi }^2}$
Rotational energy, $E=\frac{L^2}{2I}$
$\begin{array}{rl}& E=\frac{n^2{h}^2}{8{\pi }^{2}I}\cdots \cdots \cdots \cdot u\mathrm{sing}(ii)\\ & E=\frac{n^2{h}^2(m_1+m_2)}{8{\pi }^2\left(m_1{m}_2\right)r^2}\cdots \cdots \cdots \cdots \cdot u\mathrm{sing}(i)\\ & =\frac{(m_1+m_2)n^2{\hslash }^2}{2m_1{m}_2{r}^2}(\because h=\frac{h}{2\pi })\end{array}$

Summary

The effect of nucleus motion on the energy of Bohr's second postulate refines our understanding of atomic structures by considering the slight shifts in energy levels due to the nucleus's movement. This adjustment is crucial for precision in fields like spectroscopy and quantum mechanics, affecting technologies such as atomic clocks and MRI machines. The solved examples illustrate how this consideration impacts the spectral lines of positronium, deuterium, muonic atoms, and the internal energy of nuclei, highlighting the practical implications of incorporating nucleus motion in atomic models.