Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

$$Rb$$  has the configuration : $$1{s^2}2{s^2}{p^6}\,3{s^2}{p^6}{d^{10}}\,4{s^2}{p^6}5{s^1};$$
so $$n = 5,l = 0,m = 0$$     and $$s = + \frac{1}{2}$$  is correct set of quantum numbers for valence shell electron of $$Rb.$$

According to Aufbau rule
$$3s < \,3p\, < 3d\, < 4s$$

According to Aufbau principle, the orbital of lower energy $$(2s)$$  should be fully filled before the filling of orbital of higher energy starts.

Probability of finding $$1s$$  electron is maximum near the nucleus and goes on increasing till it reaches a maximum value at a distance 52.9$$\,pm$$  and then begins to decrease abruptly. Even at large distance from the nucleus, there is a finite though small probability of finding an electron of a given energy.

$$\eqalign{ & \frac{{{{\left( {KE} \right)}_1}}}{{{{\left( {KE} \right)}_2}}} = \frac{{{\nu _1} - {\nu _0}}}{{{\nu _2} - {\nu _0}}} \cr & = \frac{{\left( {1.5 \times {{10}^{15}} - 1 \times {{10}^{15}}} \right)}}{{\left( {2.0 \times {{10}^{15}} - 1 \times {{10}^{15}}} \right)}} = \frac{1}{2} \cr} $$

Electronic configuration of chlorine is $$\left[ {Ne} \right]3{s^2},3{p^5}$$
∴ Unpaired electron is found in $$3p$$  sub-shell.
$$\therefore \,\,n = 3,\,l = 1,\,m = 1$$

$$6s \to 4f \to 5d \to 6\,p\,\,{\text{for}}\,\,n = 6$$

Total energy of a revolving electron is the sum of its kinetic and potential energy.
Total energy $$ = K.E. + P.E.$$
$$\eqalign{ & = \frac{{{e^2}}}{{2r}} + \left( { - \frac{{{e^2}}}{r}} \right) \cr & = - \frac{{{e^2}}}{{2r}} \cr} $$

Wavelength of yellow light $$ = 240\,nm$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 240 \times {10^{ - 9}}m$$
Wave number $$\left( {\bar \upsilon } \right) = \frac{1}{\lambda } = \frac{1}{{240 \times {{10}^{ - 9}}m}} = 4.16 \times {10^6}{m^{ - 1}}$$

Rutherford’s scattering experiment led to the discovery of nucleus.