Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

The number of spherical nodes in any orbital $$\left( { = n - l - 1} \right)$$
For $$3p$$ -orbital, $$n = 3$$  and $$l = 1$$
∴ Number of spherical nodes
$$\eqalign{ & = n - l - 1 \cr & = 3 - 1 - 1 \cr & = 3 - 2 \cr & = 1\,node \cr} $$

$$\left( {{\text{ii}}} \right){\text{Balmer,}}\,{n_1} = 2;\,{n_2} = 3,4;$$      $${\text{region - visible}}$$
$$\left( {{\text{iv}}} \right){\text{Pfund,}}\,{n_1} = 5;\,{n_2} = 6,7;$$      $${\text{region - infrared}}$$

$$\eqalign{ & \bar v \propto \Delta E \cr & {\text{For}}\,H{\text{ - atom}} \cr & \bar v = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] \cr & {\text{For}}\,{\text{Layman}}\,{\text{series,}} \cr & \bar v\left( {\max } \right) = 13.6\left( {1 - \frac{1}{\infty }} \right) \cr & \bar v\left( {\min } \right) = 13.6\left( {1 - \frac{1}{4}} \right) \cr & \therefore \,{{\bar v}_{\max }} - {{\bar v}_{\min }} = 13.6\left( {\frac{1}{4}} \right) \cr & {\text{For Balmer series,}} \cr & \bar v\left( {\max } \right) = 13.6\left( {\frac{1}{4} - \frac{1}{\infty }} \right) \cr & \bar v\left( {\min } \right) = 13.6\left( {\frac{1}{4} - \frac{1}{9}} \right) \cr & \therefore \,\,{{\bar v}_{\max }} - {{\bar v}_{\min }} = 13.6\left( {\frac{1}{9}} \right) \cr & \frac{{\Delta {{\bar v}_{Lyman}}}}{{\Delta {{\bar v}_{Balmer}}}} = \frac{9}{4} \cr} $$

$$_{19}{K^ + },{\,_{20}}C{a^{2 + }},{\,_{21}}S{c^{3 + }}{,_{17}}C{l^\_}$$
each contains 18 electrons.

$$\eqalign{ & {r_1} = 0.529\mathop {\text{A}}\limits^{\text{o}} ;\,\,{R_{4\left( x \right)}} = {R_1} \times \frac{{{n^2}}}{Z}; \cr & {R_{4\left( x \right)}} \Rightarrow \frac{{0.529 \times {{\left( 4 \right)}^2}}}{Z};\,Z = 16 \cr} $$

The $$d$$-orbital represented by option (D) will become completely filled after gaining an electron. Therefore option (D) is correct

For Balmer series, $${n_1} = 2,{n_2} = 3,4,5,\,....$$
The spectral lines are seen in visible region.

$${\text{According to formula,}}$$     $$E = \frac{{hc}}{\lambda }\left( {v = \frac{c}{\lambda }} \right)$$
$$\eqalign{ & {\text{Energy}}\,E = h\nu \cr & {\text{3}}{\text{.03}} \times {10^{ - 19}} = \frac{{hc}}{\lambda } \cr & \lambda = \frac{{6.63 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{3.03 \times {{10}^{ - 19}}}} \cr & \,\,\,\,\, = 6.56 \times {10^{ - 7}}m \cr & \,\,\,\,\, = 6.56 \times {10^{ - 7}} \times {10^9}nm \cr & \,\,\,\,\, = 6.56 \times {10^2}nm \cr & \,\,\,\,\, = 656\,nm \cr} $$

The term spin implies that this magnetic moment is produced by the electron charge as the electron rotates about its own axis. Although this conveys a vivid mental picture of the source of the magnetism, the electron is not an extended body and its rotation is meaningless. Electron spin has no classical counterpart; the magnetic moment is a consequence of relativistic shifts in local space and time due to the high effective velocity of the electron in the atom.

$$\eqalign{ & \lambda = \frac{h}{{mv}} \cr & \,\,\,\,\, = \frac{{6.626 \times {{10}^{ - 34}}J\,s}}{{0.01\,kg \times 100\,m\,{s^{ - 1}}}} \cr & \,\,\,\,\, = 6.626 \times {10^{ - 34}}\,m \cr} $$