Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

The orbitals having the same energy energy but different in orientation, are called degenerate orbitals. $$e.g.$$  $$3d$$ - orbital, $$l = 2,\,m = - 2, - 1,0, + 1, + 2,$$      i.e. there are five different orientations represented by $${d_{xy}},{d_{yz}},{d_{zx}},{d_{{x^2} - {y^2}}}$$    and $${d_{{z^2}}}.$$

$${\text{Given,}}\,\,\Delta p = 1 \times {10^{ - 18}}g\,\,cm\,\,{s^{ - 1}}$$      $${\text{(uncertainty in momentum)}}$$
$$\eqalign{ & {\text{Mass}} = 9 \times {10^{ - 28}}g \cr & \,\,\,\,\,\Delta p = m\Delta v \cr} $$
$$1 \times {10^{ - 18}} = 9 \times {10^{ - 28}} \times \Delta v$$      $${\text{(uncertainty in velocity)}}$$
$$\Delta v = 1 \times {10^9}cm\,\,{s^{ - 1}}$$

$$mvr = \frac{{nh}}{{2\pi }} = \frac{{5h}}{{2\pi }} = \frac{{2.5h}}{\pi }$$

The sequence of energy level can be remembered by the systematic diagram as shown below

Hence, the correct order is
$$1s,2s,2p,3s,3p,4s,3d,4p,5s,$$      $$4d,5p,4f,5d,6p,7s.......$$

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

$$\eqalign{ & \bar v = R{Z^2}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) \cr & = R\left( {\frac{1}{4} - \frac{1}{9}} \right) \cr & = \frac{{5R}}{{36}} \cr} $$

$$\eqalign{ & E = \frac{{hc}}{\lambda } \times {N_A} \cr & = \frac{{6.62 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}} \times 6.02 \times {{10}^{23}}}}{\lambda } \cr & = \frac{{1.19 \times {{10}^8}}}{\lambda } \cr} $$

TIPS/Formulae :
Orbital angular momentum $$\left( {mvr} \right) = \frac{h}{{2\pi }}\sqrt {l\left( {l + 1} \right)} $$
For $$2s$$ orbital, $$l\,{\text{(azimuthal quantum number)}} = 0$$
∴ Orbital angular momentum $$ = 0.$$

$${r_n} = \frac{{52.9 \times {n^2}}}{Z}pm = {n^2}{a_0}$$     $$ \Rightarrow {a_0} = 52.9\,pm$$

$$\eqalign{ & {\text{By Heisenberg's}}\,\,{\text{uncertainty principle}} \cr & \Delta x \times \Delta {p_x} \geqslant \frac{h}{{4\pi }} \cr & {\text{or}}\,\,\Delta x \times \Delta \left( {m{v_x}} \right) \geqslant \frac{h}{{4\pi }} \cr & \Delta x \times \Delta {v_x} \geqslant \frac{h}{{4\pi m}} \cr & \Delta p = {\text{uncertainty in momentum}} \cr & \Delta x = {\text{uncertainty in position}} \cr & \Delta v = {\text{uncertainty in velocity}} \cr & m = {\text{mass of particle}} \cr & {\text{Given that,}} \cr & \Delta x = 0.1\mathop {\text{A}}\limits^{\text{o}} = 0.1 \times {10^{ - 10}}m \cr & m = 9.11 \times {10^{ - 31}}kg \cr} $$
$$h = {\text{Planck's constant}}$$     $$ = 6.626 \times {10^{ - 34}}Js$$
$$\eqalign{ & \pi = 3.14 \cr & {\text{Thus,}} \cr} $$
$$\Delta v \times 0.1 \times {10^{ - 10}}$$     $$ = \frac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 9.11 \times {{10}^{ - 31}}}}$$
$$\Delta v$$   $$ = \frac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 9.11 \times {{10}^{ - 31}} \times 0.1 \times {{10}^{ - 10}}}}m{s^{ - 1}}$$
$$\eqalign{ & \,\,\,\,\,\,\, = 5.785 \times {10^6}m{s^{ - 1}} \cr & \,\,\,\,\,\,\, = 5.79 \times {10^6}m{s^{ - 1}} \cr} $$