Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & v = \frac{{3 \times {{10}^8}}}{{10}} \cr & \,\,\,\, = 3 \times {10^7}\,m\,{s^{ - 1}} \cr & \lambda = \frac{h}{{mv}} \cr & \,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}\,J\,s}}{{9.1 \times {{10}^{ - 31}}\,kg \times 3 \times {{10}^7}\,m\,{s^{ - 1}}}} \cr & \,\,\,\,\, = 2.43 \times {10^{ - 11}}\,m \cr} $$

Difference in the energy of the energy levels involved in the transition

$$\eqalign{ & \Delta x.\,\Delta p = \frac{h}{{4\pi }}\,\,{\text{or}}\,\,\Delta x.\,m\Delta v = \frac{h}{{4\pi }}; \cr & \Delta v = \frac{{0.011}}{{100}} \times 3 \times {10^4} = 3.3\,cm{s^{ - 1}} \cr & \Delta x = \frac{{6.6 \times {{10}^{ - 27}}}}{{4 \times 3.14 \times 9.1 \times {{10}^{ - 28}} \times 3.3}} \cr & = 0.175\,cm \cr} $$

Step I : Wavelength $$\left( \lambda \right) = 589\,nm$$   $$ = 589 \times {10^{ - 9}}m$$
Frequency $$\left( \upsilon \right) = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{589 \times {{10}^{ - 9}}}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5.093 \times {10^{14}}$$      cycles per sec
Step II : Wavelength $$\left( \lambda \right) = 589.6\,nm = 589.6 \times {10^{ - 9}}m$$
$$\therefore \,\,\upsilon = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{589.6 \times {{10}^{ - 9}}}} = 5.088 \times {10^{14}}$$         cycles per sec
Energy difference between two excited states,
$$\eqalign{ & \Delta E = 6.626 \times {10^{ - 34}}\left( {5.093 - 5.088} \right){10^{14}} \cr & \,\,\,\,\,\,\,\,\,\, = 6.626 \times {10^{ - 34}} \times 5 \times {10^{ - 3}} \times {10^{14}} \cr & \,\,\,\,\,\,\,\,\,\, = 3.31 \times {10^{ - 22}}J \cr} $$

$$\eqalign{ & {\text{Let wavelength of particle be }}x \cr & {\text{So, velocity}}\left( v \right) = \frac{x}{{100}} \cr & \lambda = \frac{h}{{mv}};\,x = \frac{{h \times 100}}{{m \times x}} \cr & {x^2} = 100\frac{h}{m}\,\,{\text{or}}\,\,x = 10\sqrt {\frac{h}{m}} \cr} $$

Isoelectronic species have same no. of electrons.

ions	$${O^{2 - }}$$	$${F^ - }$$	$$N{a^ + }$$	$$M{g^{2 + }}$$
	8 + 2	9 + 1	11 - 1	12 - 2
No. of $${e^ - } = $$	10	10	10	10

Therefore $${O^{2 - }},{F^ - },N{a^ + },M{g^{ + 2}}$$     are isoelectronic

$$\eqalign{ & {\text{Total energy}} = \frac{{{\text{Potential energy}}}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{ - 5.02}}{2} \cr & {\text{Total energy}} = - 2.51\,eV \cr & {\text{Kinetic energy = }} - {\text{Total energy}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( { - 2.51\,eV} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = + \left( {2.51\,eV} \right) \cr} $$

$$r = x \times {n^2} = {3^2} \times x = 9 \times x\,pm$$

$$n = 3,\,l = 2,\,m = + 2,\,s = \pm \frac{1}{2}$$
These values of quantum numbers are possible for only one of the five $$3d$$ - orbitals as $$+2$$  value of $$m$$ is possible only for one orbital.

$$\eqalign{ & \Delta x = \pm 0.01\,\,mm = 1 \times {10^{ - 5}}\,m, \cr & m = 10\,g = 1 \times {10^{ - 2}}\,kg \cr & \Delta \nu = \frac{h}{{4\pi m \cdot \Delta x}} \cr & \,\,\,\,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}\,J\,s}}{{4 \times 3.14 \times {{10}^{ - 2}}\,kg \times {{10}^{ - 5}}\,m}} \cr & \,\,\,\,\,\,\,\, = 5.275 \times {10^{ - 28}}\,m\,{s^{ - 1}} \cr} $$