Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & E = \frac{{nhc}}{\lambda } = nhc\bar v\left( {\because \,\,\bar v = \frac{1}{\lambda }} \right) \cr & \therefore \,\,10 = nhc\bar v\,\,\,{\text{or}}\,\,\,n = \frac{{10}}{{hc\bar v}} = \frac{{10}}{{hcx}} \cr} $$

$$\eqalign{ & E = h\nu = \frac{{ch}}{\lambda };\,{\text{and}}\,\nu = \frac{c}{\lambda } \cr & 8 \times {10^{15}} = \frac{{3.0 \times {{10}^8}}}{\lambda } \cr & \therefore \,\,\lambda = \frac{{3.0 \times {{10}^8}}}{{8 \times {{10}^{15}}}} \cr & = 0.37 \times {10^{ - 7}} \cr & = 37.5 \times {10^{ - 9}}m \cr & = 4 \times {10^1}nm \cr} $$

Mass number of the element $$=81$$
i.e., $$p + n = 81$$
Let the number of protons be $$x.$$
Number of neutrons $$ = x + \frac{{31.7}}{{100}} \times x = 1.317x$$
$$\eqalign{ & \therefore \,\,x + 1.317x = 81\,\,\,{\text{or}}\,\,\,2.317x = 81 \cr & {\text{or}}\,\,\,x = \frac{{81}}{{2.317}} = 34.958 \approx 35 \cr} $$
Symbol of the element $$ = _{35}^{81}Br$$

$$\eqalign{ & \lambda = \frac{h}{{mv}} \cr & \,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}J\,s}}{{9.11 \times {{10}^{ - 31}}\,kg \times {{10}^7}\,m\,{s^{ - 1}}}} \cr & \,\,\,\,\, = 7.27 \times {10^{ - 11}}\,m \cr} $$

$$\eqalign{ & {\text{As per Bohr's}}\,{\text{postulate,}} \cr & mvr = \frac{{nh}}{{2\pi }}\,\,{\text{So}},\,\,\upsilon = \frac{{nh}}{{2\pi mr}} \cr & KE = \frac{1}{2}m{\upsilon ^2}\,{\text{So}},\,\,KE = \frac{1}{2}m{\left( {\frac{{nh}}{{2\pi mr}}} \right)^2} \cr & {\text{Since}},\,\,r = \frac{{{a_0} \times {n^2}}}{z} \cr & {\text{So,}}\,{\text{for}}\,{{\text{2}}^{{\text{nd}}}}\,{\text{Bohr}}\,{\text{orbit}} \cr & r = \frac{{{a_0} \times {2^2}}}{1} = 4{a_0} \cr & KE = \frac{1}{2}m\left( {\frac{{{2^2}{h^2}}}{{4{\pi ^2}{m^2} \times {{\left( {4{a_0}} \right)}^2}}}} \right) \cr & = \frac{{{h^2}}}{{32{\pi ^2}ma_0^2}} \cr} $$

$$\eqalign{ & \Delta E\left( {{\text{i}}{\text{.e}}.,\,{E_2} - {E_1}} \right)\,\,{\text{for}}\,\,L{i^{2 + }} = {E_2} - {E_1}\,\,{\text{is maximum as}} \cr & {E_{{n_{L{i^{2 + }}}}}} = {E_{{n_H}}} \times {Z^2} \cr} $$

Millikan devised a method known as oil drop experiment to determine the charge on the electrons.

After absorbing $$12.1\,eV$$  the electron in $$H$$ $$atom$$  is excited to 3 shell.
$$\eqalign{ & {E_n} - {E_1} = \frac{{ - 13.6}}{{{n^2}}} - \left( { - 13.6} \right) = 12.1 \cr & \frac{{ - 13.6}}{{{n^2}}} + 13.6 \cr & = 12.1 \cr & \therefore \,\,n = 3 \cr} $$
The possible transitions are = 3

According to Heisenberg uncertainty principle,
$$\Delta x \times \Delta p = \frac{h}{{4\pi }}{\text{(which is constant)}}{\text{.}}$$
As $$\Delta x$$  for electron and helium atom is same, thus momentum of electron and helium will also be same, therefore the momentum of helium atom is equal to $$5.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}.$$

The outer electronic configuration of chromium atom is $$3{d^5}4{s^1}.$$