---

We know that


Ionisation energy is the energy required to remove an electron from ground state to infinity.
Here,              

                  

                            

or,              

                        

---

Alpha-particle experiment led rutherford to the discovery of atomic nicleus.  

 
Observations of the $\mathrm{\alpha }$ particle experiment. 

i) Most of the alpha particles pass straight through the gold foil. It means that they do not suffer any collision with the gold atoms.

ii) Only about 0.14 % of incident alpha particles are scattered by more than 1^o. 

iii) About 1 alpha particle in every 8000 alpha particles deflect by more than 90^o. 

Significance of rutherford's experiment:

His experiment led him to conclude that, the entire positive charge of the atom must be concentrated in a tiny central core of the atom. The tiny massive central core of the atom was named as the atomic nucleus. 

---

Given, wavelength of the balmer series = 3646 Å 

The short limit of the Balmer series is given by the Rydberg's formula, 

              $$\begin{gathered} \overline{\mathrm{v}} = \frac{1}{\mathrm{\lambda }} =\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right) \\ \overline{\mathrm{v}} = \frac{\mathrm{R}}{4} \end{gathered}$$ 

$\therefore$         $\mathrm{R} =\frac{4}{\mathrm{\lambda }}= \frac{4}{3646}\times {10}^{10}{\mathrm{m}}^{-1}$ 

Further the wavelengths of the K_α series are given by the relation,

$\overline{\mathrm{v}} =\frac{1}{\mathrm{\lambda }} = \mathrm{R}(\mathrm{Z}-1{)}^2 \left(\frac{1}{1^2}-\frac{1}{{\mathrm{n}}^2}\right)$ 

The maximum wave number corresponds to n = ∞ and, therefore, we must have
 
                      $\overline{\mathrm{v}} = \frac{1}{\mathrm{\lambda }}= \mathrm{R}(\mathrm{Z}-1{)}^2$ 
$\Rightarrow$         $$\begin{gathered} (\mathrm{Z}-1{)}^2 = \frac{1}{\mathrm{R\lambda }} \\ =\frac{3646\times {10}^{-10}}{4\times 1\times {10}^{-10}} \\ = 911.5 \end{gathered}$$ 

$\therefore$          $(\mathrm{Z}-1) = \sqrt{911.5}$
$\Rightarrow$                   $\cong 30.2$ 

$\Rightarrow$                 $\mathrm{Z} = 31.2 \cong 31$ 

Thus, the atomic number of the element concerned is 31.

We can infer that the element having atomic number Z = 31 is Gallium.

---

Radius of nth orbit is given by, $\mathrm{r} = \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{mKZe}}^2}$

$\mathrm{i}.\mathrm{e}., \mathrm{r} \propto \frac{{\mathrm{n}}^2}{\mathrm{Z}}$ 

For hydrogen,  Z = 1, n = 1 in ground state. 

$\therefore$         $\frac{{\mathrm{n}}^2}{\mathrm{Z}} = \frac{1^2}{1} = 1$ 

For Berylium, Z = 4, as orbital radius is same,$\frac{{\mathrm{n}}^2}{\mathrm{Z}}=1$ 

$\therefore$             $$\begin{gathered} {\mathrm{n}}^2 = 1 \times \mathrm{Z} = 1\times 4 = 4 \\ \mathrm{n} = \sqrt{4} = 2 \end{gathered}$$ 

Hence n = 2.

Thus, second level of triply ionised Be⁺⁺⁺ has same radius as the ground state of hydrogen. 

Now energy of electron in nth orbit is given by, $\mathrm{E}=-\frac{2{\mathrm{\pi }}^2{\mathrm{mK}}^2{\mathrm{Z}}^2{\mathrm{e}}^4}{{\mathrm{n}}^2{\mathrm{h}}^2}$ 

$\therefore$               $\mathrm{E}\propto \frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}$

          $\frac{{\mathrm{E}}_{\left(\mathrm{Be}\right)}}{E_H} = \frac{{\left[{\mathrm{Z}}^2/{\mathrm{n}}^2\right]}_{\mathrm{Be}}}{\left[{\mathrm{Z}}^2/{\mathrm{n}}^2\right]} = \frac{16/4}{1/1} = 4$ 

Therefore, energy of ionized Be is greater than the energy of hydrogen in the ground state. 

---

The ionization energy E of a hydrogen-like Bohr atom of atomic number Z is given by, $\mathrm{E} = -{\mathrm{Rz}}^2 = -\frac{2{\mathrm{\pi k}}^2{\mathrm{z}}^2{\mathrm{me}}^4}{{\mathrm{n}}^2{\mathrm{h}}^2}$ 

where the Rydberg constant is $\mathrm{R}=\frac{2{\mathrm{\pi }}^2{\mathrm{kme}}^4}{{\mathrm{n}}^2{\mathrm{h}}^2}=2.2\times {10}^{-18}\mathrm{J}$ 

Given that, 
Ionisation energy is equal to four times the rydberg constant. 

i.e.,                $$\begin{gathered} \mathrm{E}= 4\times \mathrm{Rydberg} \mathrm{constant} \\ = 4 \mathrm{R}, \mathrm{we} \mathrm{have} \end{gathered}$$ 

$\therefore$              $4 \mathrm{R} = {\mathrm{RZ}}^2$
or,                 $\mathrm{Z} =2$ 

(i) Energy of radiation emitted (E) when the electron jumps from the first excited state to the ground state is given by,

                     $$\begin{gathered} \mathrm{E} ={\mathrm{RZ}}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ = 4\mathrm{R}\left(1-\frac{1}{4}\right) \\ = 3 \mathrm{R} \\ =3\times 2.2\times {10}^{-18}\mathrm{J} \end{gathered}$$ 

i.e.,                $\mathrm{E}=6.6 \times {10}^{-18}\mathrm{J}.$ 

Wavelength of the radiation emitted, $$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{hc}}{\mathrm{E}} \\ = \frac{6.6 \times {10}^{-34} \times 3\times {10}^8}{6.6\times {10}^{-18}} \\ = 3\times {10}^{-8}\mathrm{m} \end{gathered}$$ 

(ii) Radius of the first Bohr orbit, $$\begin{gathered} = \frac{\mathrm{Bohr} \mathrm{radius} \mathrm{of} \mathrm{hydrogen} \mathrm{atom}}{\mathrm{Z}} \\ = \frac{5\times {10}^{-11}}{2} \\ =2.5\times {10}^{-11}\mathrm{m}. \end{gathered}$$