Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{If}}\,\,n = 3, \cr & l = 0\,\,{\text{to}}\,\left( {3 - 1} \right) = 0,1,2 \cr & m = - l\,\,{\text{to}}\, + l = - 2,\, - 1,\,\,0,\, + 1,\, + 2 \cr & s = \pm \frac{1}{2} \cr} $$
Therefore, option (C) is not a permissible set of quantum numbers.

By Heisenberg’s uncertainty principle
$$\Delta x \times \Delta p \geqslant \frac{h}{{4\pi }}$$
when the position of electron and helium atom is same and momentum of electron is known within a range, therefore the momentum of helium atom is also equal to the momentum of electron, i.e.
$$5 \times {10^{ - 26}}kg\,m\,{s^{ - 1}}$$

$$\eqalign{ & {\text{From Bohr's theory}}, \cr & \nu = 2.18 \times {10^6} \times \frac{Z}{n}m/s \cr & \,\,\,\,\, = 2.18 \times {10^6} \times \frac{1}{1} \cr & \,\,\,\,\, = 2.18 \times {10^6}\,m/s \cr} $$

Charge carried by one electron $$ = 1.6022 \times {10^{ - 19}}C$$
Electrons present in $$5.5 \times {10^{ - 16}}\,C$$
$$\eqalign{ & = \frac{{5.5 \times {{10}^{ - 16}}}}{{1.6022 \times {{10}^{ - 19}}}} \cr & = 3432 \cr} $$

$$\lambda = 242\,nm = 242 \times {10^{ - 9}}\,m$$
Energy required to ionise one atom of $$Na,E = \frac{{hc}}{\lambda }$$
$$\eqalign{ & = \frac{{\left( {6.626 \times {{10}^{ - 34}}\,J\,s} \right) \times \left( {3 \times {{10}^8}\,m\,{s^{ - 1}}} \right)}}{{242 \times {{10}^{ - 9}}m}} \cr & = 8.214 \times {10^{ - 19}}\,J/atom \cr} $$

For $$g$$  orbital, value of $$l$$  is 4.
Since $$l=n-1,n$$    should be 5.

$$\eqalign{ & \frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] \cr & \,\,\,\,\,\, = 109677\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right]c{m^{ - 1}} \cr & \,\,\,\,\,\, = 20564.4\,\,c{m^{ - 1}} \cr & \lambda = \frac{I}{{20564.4\,\,c{m^{ - 1}}}} \cr & \,\,\,\,\, = 486 \times {10^{ - 7}}\,cm\,\,\,{\text{or}}\,\,\,486 \times {10^{ - 9}}\,m \cr} $$
Colour corresponding to this wavelength is blue.

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

$$\eqalign{ & E = \frac{{hc}}{\lambda };\,{\lambda _1} = 2000\,\mathop {\text{A}}\limits^{\text{o}} ;{\lambda _2} = 4000\mathop {\text{A}}\limits^{\text{o}} ; \cr & {\text{so}}\,\,\frac{{{E_1}}}{{{E_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{4000}}{{2000}} = 2 \cr} $$

Rutherford's experiment was actually $$\alpha - $$ particle scattering experiment. $$\alpha - $$ Particle is doubly positively charged helium ion i.e., $$He$$ - nucleus.