Total no. of electrons $$ = 2{n^2}$$
if $$n = 4,$$  no. of electrons $$=32$$
Total number of electrons with $${m_s} = \frac{{ - 1}}{2}$$    will be 16.

NOTE : Exactly half filled orbitals are more stable than nearly half filled orbitals.
$$Cr\left( {At.\,\,no.\,\,24} \right)$$    has configuration $$\left[ {Ar} \right]3{d^5},4{s^1}.$$

$$\begin{array}{l}{}_{18}X=1s^2,2s^2,2p^6,\underset{}{\underbracket{3s^2,3p^6}}\\ {}_{19}Y=1s^2,2s^2,2p^6,\underset{}{\underbracket{3s^2,3p^6}},4s^1\\ {}_{20}Z=1s^2,2s^2,2p^6,\underset{}{\underbracket{3s^2,3p^6}},4s^2\end{array}$$

TIPS/Formulae :
$$\eqalign{ & \Delta x.\Delta p = \frac{h}{{4\pi }};\,\,\,{\text{or}}\,\,\Delta x.m.\Delta v = \frac{h}{{4\pi }} \cr & \therefore \,\Delta v = \frac{{6.62 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 0.025 \times {{10}^{ - 5}}}} \cr & = 2.1 \times {10^{ - 28}}m{s^{ - 1}} \cr} $$

In $$_{38}^{88}Sr,$$
Atomic number = No. of protons = No. of electrons = 38
Atomic mass = 88
Number of neutrons = 88 - 38 = 50

Energy of subshells follow the order $$1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},4{s^2},3{d^{10}},4{p^6},5s,4d,5p,6s.....$$           we can fill them by electrons upto $$4p$$  hence 36 electrons.

$$\eqalign{ & {r_n} = {a_0}{n^2} \cr & r = {a_0} \times {\left( 3 \right)^2} = 9{a_0} \cr & mvr = \frac{{nh}}{{2\pi }}; \cr & mv = \frac{{nh}}{{2\pi r}} = \frac{{3h}}{{2\pi \times 9{a_0}}} = \frac{h}{{6\pi {a_0}}} \cr & \lambda = \frac{h}{{mv}} = \frac{h}{h} \times 6\pi {a_0} = 6\pi {a_0} \cr} $$

According to Heisenberg uncertainty principle.
$$\eqalign{ & \Delta x.m\Delta v = \frac{h}{{4\pi }} \cr & \Delta x = \frac{h}{{4\pi m\Delta v}} \cr & {\text{Here}}\,\,\Delta v = \frac{{600 \times 0.005}}{{100}} = 0.03 \cr & {\text{So}},\,\,\Delta x = \frac{{6.6 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 0.03}} \cr & = 1.92 \times {10^{ - 3}} \cr} $$

$${\text{Angular momentum}}$$     $$ = m\nu r = \frac{{nh}}{{2\pi }};\nu = \frac{{nh}}{{2\pi mr}}$$
$$\eqalign{ & {\text{As, Time period}} = \frac{{{\text{Circumference}}}}{{{\text{Velocity}}}} \cr & T = \frac{{2\pi r}}{\nu } = \frac{{2\pi r}}{{\frac{{nh}}{{2\pi mr}}}} = \frac{{4{\pi ^2}m{r^2}}}{{nh}} \cr} $$

As a very few $$\alpha $$ - particles were deflected by sharp angles, it was concluded that the positive charge is concentrated in a very small volume of the atom.