$$\eqalign{ & \Delta \nu \cdot \Delta x = \frac{h}{{4\pi m}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}J\,s}}{{4 \times 3.14 \times {{10}^{ - 6}}\,kg}} \cr} $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.52 \times {10^{ - 28}}\,{m^2}\,{s^{ - 1}}$$        $$\left( {1\,\,mg = {{10}^{ - 6}}\,kg} \right)$$

One $$p-$$orbital can accommodate up to two electrons with opposite spin while $$p-$$subshell can accommodate upto six electrons.

$$\eqalign{ & Z = 3\,\,{\text{for}}\,L{i^{2 + }}\,ions \cr & {\text{So}}\,\,\,{r_n} = \frac{{52.9 \times {n^2}}}{Z}pm \cr & n = 3,Z = 3 \cr & {r_n} = \frac{{52.9 \times {{\left( 3 \right)}^2}}}{3}pm \cr & = 158.7\,pm \cr & {\text{Also, linear momentum}}\,\left( {mv} \right) \cr & = 7.3 \times {10^{ - 34}}kg\,m{s^{ - 1}} \cr & {\text{Then angular momentum will be}} \cr & \omega = \left( {mv} \right) \times r \cr & = \left( {7.3 \times {{10}^{ - 34}}kg\,m{s^{ - 1}}} \right)\left( {158.7\,pm} \right) \cr & = 7.3 \times {10^{ - 34}}kg\,m{s^{ - 1}} \times \left( {158.7 \times {{10}^{ - 12}}m} \right) \cr & = 11.58 \times {10^{ - 48}}kg\,{m^2}{s^{ - 1}} \cr} $$

As per Pauli Exclusion Principle "no two electrons in the same atom can have all the four quantum numbers equal or an orbital cannot contain more than two electrons and it can accommodate two electrons only when their directions of spins are opposite".

The light emitted by a sample ( excited hydrogen atom or any other element ), which is passed through a prism and then separated into certain discrete wavelength is called atomic emission spectrum.

By Heisenberg uncertainty Principle
$$\Delta x \times \Delta p = \frac{h}{{4\pi }}$$     ( which is constant )
As $$\Delta x$$  for electron and helium atom is same thus momentum of electron and helium will also be same therefore the momentum of helium atom is equal to $$5 \times {10^{ - 26}}kg.m.{s^{ - 1}}$$

The electron will enter into an orbital with minimum value of $$n + l.$$

If two orbitals have same value for $$\left( {n + l} \right),$$  the orbital with lower value of $$n$$ will have lower energy.

$$\Delta x \cdot m\Delta v = \frac{h}{{4\pi }}$$
$${10^{ - 8}}\,m \times m \times 5.26 \times {10^{ - 25}}\,m\,{s^{ - 1}}$$       $$ = \frac{{6.626 \times {{10}^{ - 34}}\,J\,s}}{{4 \times 3.14}}$$
$$\eqalign{ & m = \frac{{6.626 \times {{10}^{ - 34}}\,J\,s}}{{4 \times 3.14 \times {{10}^{ - 8}}\,m \times 5.26 \times {{10}^{ - 25}}\,m\,{s^{ - 1}}}} \cr & \,\,\,\,\,\, = 0.01\,kg \cr} $$

From the expression of Bohr’s theory, we know that
$$\eqalign{ & {m_e}{v_1}{r_1} = {n_1}\frac{h}{{2\pi }} \cr & \& \,\,{m_e}{v_2}{r_2} = {n_2}\frac{h}{{2\pi }} \cr & \frac{{{m_e}{v_1}{r_1}}}{{{m_e}{v_2}{r_2}}} = \frac{{{n_1}}}{{{n_2}}}\frac{h}{{2\pi }} \times \frac{{2\pi }}{h} \cr & {\text{Given,}}\,\,{r_1} = 5{r_2},{n_1} = 5,{n_2} = 4 \cr & \frac{{{m_e} \times {v_1} \times 5{r_2}}}{{{m_e} \times {v_2} \times {r_2}}} = \frac{5}{4} \cr & \Rightarrow \frac{{{v_1}}}{{{v_2}}} = \frac{5}{{4 \times 5}} = \frac{1}{4} = 1:4 \cr} $$