If we assume the atom to be hydrogen like, energy of $${n^{th}}$$  energy level
$${E_n} = - \frac{{{E_1}}}{{{n^2}}}$$
where $${{E_1}}$$ is energy of first energy level
$$\eqalign{ & {E_2} = - \frac{{{E_1}}}{{{2^2}}} \cr & = - \frac{{{E_1}}}{4} \cr & = \frac{{ - 21.79 \times {{10}^{ - 12}}}}{4} \cr & = - 5.447 \times {10^{ - 12}}erg\,\,per\,\,atom. \cr} $$

$$\eqalign{ & E = h\nu \cr & {\text{and}}\,\,\nu = \left( {\frac{c}{\lambda }} \right) \cr & {\nu _a} = {10^{15}},{\nu _b} = {10^{14}}, \cr & {\nu _c} = {10^{17}},{\nu _d} = 0.85 \times {10^{15}} \cr & {\text{and}}\,\,{\nu _e} = 3 \times {10^{15}}. \cr} $$

The species $$CO,N{O^ + },C{N^ - }$$   and $$C_2^{2 - }$$ contain 14 electrons each.

Only a few $$\alpha $$ - particles ( 1 in 20,000 ) bounced back.

According to Hund’s rule of maximum multiplicity, An orbital can accommodate a maximum number of 2 electrons of exactly opposite spin. Hence, option (A) is correct.
Caution Remember, maximum number of electrons in an orbital do not depend upon the quantum numbers as given in the question.

Bohr's model of an atom describes orbit as a clearly defined path and this could be only possible when both position and the velocity of the electron are known exactly at the same time. This is not possible according to Heisenberg uncertainty principle.

TIPS/Formulae : The following is the increasing order of wavelength or decreasing order of energy of electromagnetic radiations :

Among given choices radiowaves have maximum wavelength.

The value of $$l$$ varies from 0 to $$(n - 1)$$  and the value of $$m$$ varies from $$- l$$  to $$+l$$  through zero.
The value of $$'s'\,\, \pm \,\,\frac{1}{2}$$  which signifies the spin of electron. The correct sets of quantum number are following:

	$$n$$	$$l$$	$$m$$	$$s$$
(ii)	2	1	1	$$ - \frac{1}{2}$$
(iv)	1	0	0	$$ - \frac{1}{2}$$
(v)	3	2	2	$$ + \frac{1}{2}$$

According to Heisenberg's uncertainty principle
$$\eqalign{ & \Delta p \times \Delta x \geqslant \frac{h}{{4\pi }} \cr & {\text{Uncertainty in momentum}} \cr & \Delta p = 1 \times {10^{ - 5}}kg\,m/s \cr & 1 \times {10^{ - 5}} \times \Delta x = \frac{{6.62 \times {{10}^{ - 34}}}}{{4 \times \frac{{22}}{7}}}\,\,\left( {{\text{Given}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta x = \frac{{6.62 \times {{10}^{ - 34}} \times 7}}{{1 \times {{10}^{ - 5}} \times 4 \times 22}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5.265 \times {10^{ - 30}}m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \approx 5.27 \times {10^{ - 30}}m \cr} $$

The possible quantum numbers for $$4f$$ electron are
$$\eqalign{ & n = 4,\, \cr & l = 3,\, \cr & m = - 3,\, - 2 - 1,\,\,0,\,\,1,\,\,2,\,\,3 \cr & s = \pm \frac{1}{2} \cr} $$