Only overall neutrality of an atom could be explained correctly by Thomson model of atom.

Paschen, $${n_1} = 3;{n_2} = 4,5\,......;$$     region-infrared.

The electronic configuration of

Hence, the number of unpaired electron in $${N^{2 + }}$$  is 1.

$$\eqalign{ & \lambda = \frac{h}{{mv}} \cr & \,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}\,J\,s}}{{9.11 \times {{10}^{ - 31}}\,kg \times 2.19 \times {{10}^6}\,m\,{s^{ - 1}}}} \cr & \,\,\,\,\, = 3.32 \times {10^{ - 10}}\,m \cr} $$

Angular momentum of an electron in nth orbital is given by,
$$\eqalign{ & mvr = \frac{{nh}}{{2\pi }} \cr & {\text{For}}\,n = 5,\,{\text{we}}\,{\text{have}} \cr & {\text{Angular momentum of electron}} \cr & = \frac{{5h}}{{2\pi }} = \frac{{2.5h}}{\pi } \cr} $$

Total radiations are = 6

$$n = 3,l = 2$$   means $$3d$$ orbital

i.e. in an atom only one orbital can have the value $${m_1} = + 2$$

$$\eqalign{ & {\nu _1} = \frac{c}{{{\lambda _1}}} \cr & {\nu _2} = \frac{c}{{{\lambda _2}}} = \frac{c}{{3{\lambda _1}}} \cr & \% \,{\text{change in frequency}} \cr & = \frac{{{\nu _2} - {\nu _1}}}{{{\nu _1}}} \times 100 \cr & = \frac{{\frac{c}{{3{\lambda _1}}} - \frac{c}{{{\lambda _1}}}}}{{\frac{c}{{{\lambda _1}}}}} \times 100 \cr & = \frac{{ - \frac{{2c}}{{3{\lambda _1}}} \times 100}}{{\frac{c}{{{\lambda _1}}}}} \cr & = - 66\% \cr} $$

$${X^{3 - }}$$  has 14 protons, i.e., $$X$$ also has 14 protons and therefore 14 electrons
$$\eqalign{ & X = 14 = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^2} \cr & {X^{2 + }} = 1{s^2},2{s^2}2{p^6},3{s^2} = 12\,{\text{electrons}} \cr} $$

$$\eqalign{ & {\text{Atomic number}} = 24 \cr & {\text{Electronic configuration is}}\,\,3{d^5},4{s^1} \cr & {\text{No}}{\text{. of valence}}\,\,{e^{ - 1}}\left( s \right) = 6 \cr & {\text{Total spin}}\, = \pm \frac{1}{2} \times 6 = \pm 3 \cr & {\text{Magnetic moment}} \cr & = \sqrt {n\left( {n + 2} \right)} \cr & = \sqrt {6\left( {6 + 2} \right)} \cr & = \sqrt {48} \,BM \cr} $$