Angular momentum $$ = mrv = J$$
$$\therefore v = \frac{J}{{mr}}$$
$$K.E.$$  of electron $$ = \frac{1}{2}m{v^2} = \frac{1}{2}m{\left( {\frac{J}{{mr}}} \right)^2} = \frac{{{J^2}}}{{2m{r^2}}}$$

Speed of electron in n^th orbit
$$\eqalign{ & {V_n} = \frac{{2\pi \,KZ{e^2}}}{{nh}} \cr & V = \left( {2.19 \times {{10}^6}m/s} \right)\frac{Z}{n} \cr & V = \left( {2.19 \times {{10}^6}} \right)\frac{2}{3}\left( {Z = 2\,\& \,n = 3} \right) \cr & V = 1.46 \times {10^6}m/s \cr} $$

The maximum number of electrons in an orbit is $$2{n^2}.n > 4$$   is not allowed.
Therefore the number of maximum electron that can be in first four orbits are
$$\eqalign{ & 2{\left( 1 \right)^2} + 2{\left( 2 \right)^2} + 2{\left( 3 \right)^2} + 2{\left( 4 \right)^2} \cr & = 2 + 8 + 18 + 32 = 60 \cr} $$
Therefore, possible element are 60.

KEY CONCEPT : $$E = Rhc\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$
$$E$$ will be maximum for the transition for which $$\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$   is maximum. Here $${n_2}$$ is the higher energy level.
Clearly, $$\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$   is maximum for the third transition, i.e. $$2 \to 1.$$  I transition represents the absorption of energy.

According to question, for wavelength $${\lambda _1},$$
$${n_1} = 4\,\,{\text{and}}\,\,{n_2} = 3$$
and for $${\lambda _2},{n_1} = 3\,\,{\text{and}}\,\,{n_2} = 2$$
and we know that, $$\frac{{hc}}{\lambda } = - 13.6\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$
$$\eqalign{ & {\text{So,}}\,\,{\text{for}}\,\,{\lambda _1} \cr & \Rightarrow \frac{{hc}}{{{\lambda _1}}} = - 13.6\left[ {\frac{1}{{{{\left( 4 \right)}^2}}} - \frac{1}{{{{\left( 3 \right)}^2}}}} \right] \cr & \frac{{hc}}{{{\lambda _1}}} = 13.6\left[ {\frac{7}{{144}}} \right]\,.......\left( {\text{i}} \right) \cr} $$
Similarly, for $${{\lambda _2}}$$
$$\eqalign{ & \Rightarrow \frac{{hc}}{{{\lambda _2}}} = - 13.6\left[ {\frac{1}{{{{\left( 3 \right)}^2}}} - \frac{1}{{{{\left( 2 \right)}^2}}}} \right] \cr & \frac{{hc}}{{{\lambda _2}}} = 13.6\left[ {\frac{5}{{36}}} \right]\,.......\left( {{\text{ii}}} \right) \cr} $$
Hence, from Eqs. (i) and (ii), we get
$$\frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{20}}{7}$$

The possible number of the spectral lines is given
$$ = \frac{{n\left( {n - 1} \right)}}{2} = \frac{{4\left( {4 - 1} \right)}}{2} = 6$$

As $$F = \frac{{m{v^2}}}{r} = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\beta }{{{r^3}}}} \right)$$
and $$mvr = \frac{{nh}}{{2\pi }}$$
$$ \Rightarrow v = \frac{{nh}}{{2\pi mr}}$$
$$\eqalign{ & \therefore m{\left( {\frac{{nh}}{{2\pi mr}}} \right)^2} \times \frac{1}{r} = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\beta }{{{r^3}}}} \right) \cr & {\text{or,}}\,\,\frac{1}{{{r^2}}} + \frac{\beta }{{{r^3}}} = \frac{{m{n^2}{h^2}4\pi { \in _0}}}{{4{\pi ^2}{m^2}{e^2}{r^3}}} \cr & {\text{or,}}\,\,\frac{{{a_0}{n^2}}}{{{r^3}}} = \frac{1}{{{r^2}}} + \frac{\beta }{{{r^3}}}\,\,\left( {\because {a_0} = \frac{{{ \in _0}{h^2}}}{{m\pi {e^2}}}\,\,{\text{Given}}} \right) \cr & \therefore r = {a_0}{n^2} - \beta \cr} $$

Energy of excitation,
$$\eqalign{ & \Delta E = 13.6\,{\pi ^2}\left( {\frac{1}{{{\eta _1}}} - \frac{1}{{{\eta _2}}}} \right)eV \cr & \Rightarrow \Delta E = 13.6{\left( 3 \right)^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = 108.8\,eV \cr} $$

When continuous light from a source is examined directly in a spectroscope, we observe the emission spectrum of the source. The sodium vapour spectrum, consists of a few isolated bright lines. Each bright line corresponds to a particular wavelength. It is emitted by the atoms in the gaseous state.
When continuous light from a source is made to pass through an absorbing substance and then examined in a spectroscope, we observe absorption spectrum of the substance.
A band spectrum is emitted by chemical compounds in the vapour state. It is therefore a molecule spectrum.
A continuous emission spectrum consists of a wide range of unseparated wavelengths.

No. of particles scattered through an angle
$$\eqalign{ & \theta = N\left( \theta \right) = \frac{{k\,{{\rlap{-} Z}^2}}}{{{{\sin }^4}\left( {\frac{\theta }{2}} \right){{\left( {K.E.} \right)}^2}}} \cr & \therefore 28 = \frac{{4kc{z^2}}}{{{{\left( {K.E.} \right)}^2}}}\,{\text{for}}\,\theta = {90^ \circ } \cr & \therefore \frac{{k{z^2}}}{{{{\left( {K.E.} \right)}^2}}} = \frac{{28}}{4} = 7 \cr & \therefore N\left( {{{60}^ \circ }} \right) = \frac{7}{{{{\sin }^4}\left( {\frac{{{{60}^ \circ }}}{2}} \right)}} = 16 \times 7 = 112/\min . \cr & N\left( {{{120}^ \circ }} \right) = \frac{7}{{{{\sin }^4}\left( {\frac{{{{120}^ \circ }}}{2}} \right)}} = 12.4/\min \cr} $$