Effect Of Nucleus Motion On Energy MCQ - Practice Questions with Answers

Effect of Nucleus motion on Energy-

Till now in Bohr's model we have assumed that the 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 compare to the mass of the nucleus so all the mass is assumed to be concetrated at the center of the nucleus. But actually, centre of mass of nucleaus-electron system is close to nucleaus as it is heavy and to keep the centre of mass at rest, both electron and nucleaus revolve around their centre of mass like a double star system as shown in figure. If r is the distance of electron from nucleaus, the distances of

nucleaus 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 in the atom, nucleaus 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 above system, we can analyze the motion of electron with respect to nucleaus by assuming nucleus to be at rest and the mass of electron replaced by its reduced mass ${\mu }_{\mathrm{e},\text{ given as - }}$

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

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 by its reduced mass, the equation become -

$$r_n^{\prime }=\frac{n^2{h}^2}{4{\pi }^{2}kZ{e}^2{\mu }_e}\Rightarrow r_n^{\prime }=\frac{n^2{h}^2(m_N+m_e)}{4{\pi }^{2}kZ{e}^2{m}_e{m}_N}$$

or $r_n^{\prime }=r_n\times \frac{m_e}{{\mu }_e}\Rightarrow r=(0.529\mathrm{A})\frac{m{n}^2}{\mu Z}$
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 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 -

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

Thus, we can say that the energy of electron 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 nucleaus.