Effect Of Nucleus Motion On Energy - Practice Questions & MCQ

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 much small and negligible as compared to the mass of the nucleus so all the mass is assumed to be concentrated at the center of the nucleus. But actually, centre of mass of the nucleus-electron system is close to nucleus as it is heavy, and to keep the centre of mass at rest, both electron and nucleus revolve around their centre of mass like a double star system as shown in figure. If r is the distance of electron from nucleus, the distances of 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 $\quad r_2=\frac{m_N r}{m_N+m_e}$

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 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 k Z e^2 m_e}
$

Now after replacing the electron mass with its reduced mass, the equation becomes-
or $\quad r_n^{\prime}=r_n \times \frac{m_e}{\mu_e} \Rightarrow r=(0.529 \mathrm{~A}) \frac{m n^2}{\mu Z}$

$
r_n^{\prime}=\frac{n^2 h^2}{4 \pi^2 k Z e^2 \mu_e} \Rightarrow r_n^{\prime}=\frac{n^2 h^2\left(m_N+m_e\right)}{4 \pi^2 k Z e^2 m_e m_N}
$

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

$
v_n=\frac{2 \pi k Z e^2}{n h}
$
 

Similarly for the 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{aligned}
& E_n^{\prime}=-\frac{2 \pi^2 k^2 Z^2 e^4 m_N m_e}{n^2 h^2\left(m_N+m_e\right)} \\
& E_n^{\prime}=E_n \times \frac{\mu_e}{m_e} \Rightarrow E_n=-(13.6 \mathrm{eV}) \frac{Z^2}{n^2}\left(\frac{\mu}{m}\right)
\end{aligned}
$

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 motion of nucleus.