The electronic configuration for phosphorus (15) is $$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^3}.$$

Possible values of $$l$$ and m depend upon the value of $$n$$
$$l = 0\,{\text{to}}\,\left( {n - 1} \right)$$
$$m = - l\,{\text{to}}\, + l$$    through zero
$$s = + \frac{1}{2}\,{\text{and}}\, - \frac{1}{2}$$
Thus for $$n = 3,$$
$$l$$ may be 0, 1 or 2; but not 3
$$m$$ may be –2, –1, 0, +1 or +2
$$s$$ may be $$ + \frac{1}{2}\,\,{\text{or}}\,\, - \frac{1}{2}$$

$$\eqalign{ & {\text{Quantum efficiency}} \cr & = \frac{{{\text{Number of moles of substance reacted}}}}{{{\text{Number of Quanta absorbed}}}} \cr & = \frac{{3 \times {{10}^{ - 3}} \times 6 \times {{10}^{23}}}}{{3 \times {{10}^{18}} \times 60 \times 10}} \cr & = 1 \cr} $$

$${\text{We know that}}\,\,{r_n}\left( {H{\text{ - like}}} \right)$$      $$ = \frac{{{r_n}\left( {H{\text{ - }}atom} \right) \times {n^2}}}{Z}$$
$$\eqalign{ & {\text{For ground state,}}\,\,n = 1 \cr & \therefore \,\,{r_n}\left( {L{i^{2 + }}} \right) = \frac{{0.53\mathop {\text{A}}\limits^{\text{o}} \times {{\left( 1 \right)}^2}}}{3} \cr & \left( {Li,Z = 3} \right) = 0.17\,\mathop {\text{A}}\limits^{\text{o}} \cr} $$

TIPS/Formulae :
Number of radial nodes $$ = \left( {n - l - 1} \right)$$
$$\eqalign{ & {\text{For}}\,3s\,:\,\,n = 3,\,\,l = 0\left( {{\text{Number}}\,{\text{of}}\,{\text{radial}}\,{\text{node}} = 2} \right) \cr & {\text{For}}\,2p\,:\,\,n = 2,\,\,l = 1\left( {{\text{Number}}\,{\text{of}}\,{\text{redial}}\,{\text{node}} = 0} \right) \cr} $$

$$\eqalign{ & _{11}Na = 1{s^2}2{s^2}2{p^6}3{s^1} \cr & {\text{For}}\,\,3{s^1},n = 3,l = 0,{m_l} = 0,s = - \frac{1}{2} \cr} $$

$$\eqalign{ & E = {E_1} + {E_2};\,\,\frac{{hc}}{\lambda } = \frac{{hc}}{{{\lambda _1}}} + \frac{{hc}}{{{\lambda _2}}} \cr & \Rightarrow \frac{{hc}}{\lambda } = hc\left( {\frac{{{\lambda _2} + {\lambda _1}}}{{{\lambda _1}{\lambda _2}}}} \right);\,\lambda = \frac{{{\lambda _1}{\lambda _2}}}{{{\lambda _{`1}} + {\lambda _2}}} \cr} $$

The orientation of an atomic orbital is governed by magnetic quantum number.

(A) According to de-Broglie’s equation,
$${\text{Wavelength}}\left( \lambda \right) = \frac{h}{{mv}}$$
where, $$h =$$  Planck's constant.
Thus, statement (A) is correct.
(B) According to Heisenberg uncertainty principle, the uncertainties of position $$\left( {\Delta x} \right)$$  and momentum $$\left( {p = m\Delta v} \right)$$   are related as
$$\eqalign{ & \Delta x.\,\,\Delta p \geqslant \frac{h}{{4\pi }} \cr & {\text{or,}}\,\,\Delta x.\,\,m\Delta v \geqslant \frac{h}{{4\pi }} \cr & \Delta x.\,\,m.\,\,\Delta a.\,\,\Delta t \geqslant \frac{h}{{4\pi }} \cr & \left[ {\frac{{\Delta v}}{{\Delta t}} = \Delta a,\,\,a = {\text{acceleration }}} \right] \cr & {\text{or,}}\Delta {\text{x}}\,.\,\,F.\,\,\Delta t \geqslant \frac{h}{{4\pi }}\,\,\left[ {\because \,\,F = m.\,\,\Delta a} \right] \cr & {\text{or,}}\,\,\Delta {\text{E}}{\text{.}}\,\,\Delta {\text{t}} \geqslant \frac{h}{{4\pi }}\,\,\left[ {\because \Delta E = F.\,\,\Delta x.\,\,E = {\text{energy}}} \right] \cr} $$
Thus, statement (B) is correct.
(C) The half and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement. Thus statement (C) is correct.
(D) For a single electronic species like $$H,$$ energy depends on value of $$n$$  and does not depend on $$l$$ Hence energy of $$2s$$ - orbital. and $$2p$$ - orbital is equal in case of hydrogen like species. Therefore, statement (D) is incorrect.

Energy of one photon is given by the expression,
$$\eqalign{ & E = h\upsilon \cr & E = 6.626 \times {10^{ - 34}}J\,\,s \times 5 \times {10^{14}}{s^{ - 1}} \cr & \,\,\,\,\,\, = 3.313 \times {10^{ - 19}}J \cr} $$