Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\Delta x \cdot m\Delta v = \frac{h}{{4\pi }}$$
$$\Delta x = 1000\,\mathop {\text{A}}\limits^{\text{o}} = 1000 \times {10^{ - 10}}\,m\,\,\,$$      $${\text{or}}\,\,\,{10^{ - 7}}\,m$$
$$\eqalign{ & \Delta v = \frac{{6.626 \times {{10}^{ - 34}}\,J\,s}}{{4 \times 3.14 \times 9.1 \times {{10}^{ - 31}}\,kg \times {{10}^{ - 7}}\,m}} \cr & \,\,\,\,\,\,\,\, = 5.79 \times {10^2}\,m/s \cr} $$

The spectrum of an atom depends on the number of electrons present in it. Here, helium has two electrons, so the spectrum of $$L{i^ + }\left( {Z = 3} \right)$$   is similar to that of helium because both He and $$L{i^ + }$$  have two electrons.

$$\eqalign{ & \frac{e}{m}\,{\text{for}}\,{\text{neutron}} = \frac{0}{1} = 0;\,\, \cr & \alpha {\text{ - particle}} = \frac{2}{4} = 0.5; \cr & {\text{proton}} = \frac{1}{1} = 1\,;\,{\text{electron}} = \frac{1}{{\frac{1}{{1837}}}} = 1837 \cr} $$

$$\eqalign{ & {\text{Total energy of third shell}} \cr & = \frac{{ - 13.6}}{{{3^2}}} \cr & = - 1.51\,eV \cr & K.E. = - {\text{Total energy}} \Rightarrow 1.51\,eV \cr & P.E. = 2 \times T.E. = - 3.02\,eV \cr} $$

$$\eqalign{ & {\text{According to de - Broglie relation,}} \cr & \lambda = \frac{h}{{mv}} = \frac{h}{p} \cr & {\text{where,}}\,\,\lambda = {\text{wavelength}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,h = {\text{Planck's}}\,{\text{constant}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p = {\text{momentum}} \cr & {\text{Here,}}\,\,h = 6.625 \times {10^{ - 34}}J\,s \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\lambda = {10^{ - 17}}m \cr & \therefore \,\,p = \frac{h}{\lambda } \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{6.625 \times {{10}^{ - 34}}}}{{{{10}^{ - 17}}}} \cr & \,\,\,\,\,\,\,\,\,\, = 6.625 \times {10^{ - 34}} \times {10^{17}} \cr & \,\,\,\,\,\,\,\,\,\, = 6.625 \times {10^{ - 17}}kg\,m\,{s^{ - 1}} \cr} $$

As packet of energy equal to $$hv$$ ; as wave having frequency $$v.$$

Spatial orientation of the orbital with respect to standard set of cordinate axis. Magnetic quantum number $$-1$$  is possible only when the azimuthal quantum number have value $$l = 1,$$  which is possible for $$p,$$ $$d$$ and $$f$$ - subshells but not for $$s$$ - subshell because the value of $$l$$ for $$s$$ - subshell is zero.

According to Heisenberg’s uncertainty principle
$$\Delta x.\Delta p = \frac{h}{{4\pi }}$$
$${\text{Given,}}\,\,\Delta x = \Delta p$$     $$\left( {\Delta x = {\text{uncertainty in position}}} \right)$$
$$\eqalign{ & \,\,\,\,\,\,\,\,{\left( {\Delta p} \right)^2} = \frac{h}{{4\pi }}\,\,\left( {\Delta p = m \times \Delta v} \right) \cr & \,\,\,\,\,{m^2}\Delta {v^2} = \frac{h}{{4\pi }}\,\,\,\left( {m = {\text{mass}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\Delta {v^2} = \frac{h}{{{m^2}4\pi }} \cr} $$
    $$ \Rightarrow \,\,\Delta v = \frac{1}{{2m}}\sqrt {\frac{h}{\pi }} $$     $$\left( {\Delta v = {\text{uncertainty in velocity}}} \right)$$

$$m = - l$$   to $$+l,$$  through zero thus for $$l$$ = 2, values of $$m $$  will be -2, -1, 0, +1, +2.
Therefore for $$l$$ = 2, $$m$$ cannot have the value –3.

$$\eqalign{ & \Delta E = \frac{{hc}}{\lambda } \cr & \lambda = \frac{{6.6 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}\,m\,{s^{ - 1}}}}{{3 \times {{10}^{ - 19}}\,J}} \cr & \,\,\,\,\, = 6.6 \times {10^{ - 7}}\,m \cr} $$