Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

Cathode rays are not electromagnetic wave because they do not have electric and magnetic components perpendicular to each other.

$$\eqalign{ & {\text{Energy of an electron in an orbit,}} \cr & {E_n} = - \frac{{2{\pi ^2}m{e^4}{Z^2}}}{{{n^2}{h^2}}} \cr & \,\,\,\,\,\,\,\, = - \frac{{13118\,\,{Z^2}}}{{{n^2}}}kJ\,mo{l^{ - 1}} \cr & {E_n} = - \frac{{21.8 \times {{10}^{ - 12}}\,{Z^2}}}{{{n^2}}}erg\,\,ato{m^{ - 1}} \cr & {E_n} = - \frac{{21.8 \times {{10}^{ - 19}}\,{Z^2}}}{{{n^2}}}J\,ato{m^{ - 1}} \cr & {E_n} = - \frac{{13.6\,\,{Z^2}}}{{{n^2}}}eV\,ato{m^{ - 1}} \cr & \,\,\,\,\,\,\,\, = \frac{{ - 13.6}}{{{n^2}}}eV\,atom \cr} $$
( $${\because \,\,Z = 1}$$   atomic number for hydrogen atom )

$$\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{{{n_1}}}{{{n_2}}}} \right)^2} = \frac{4}{9}$$

$${\text{Change in }}P.E. = - \frac{{2x}}{4} + \left( {2x} \right) = \frac{3}{2}x$$

$$l = 2$$  represent $$d$$ orbital for which

The number of electrons ejected is directly proportional to the intensity of light.

The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.

The principal quantum number $$(n)$$ is related to the size of the orbital ( $$n$$ = 1, 2, 3 . . . . . )

$$\left( {\Delta E} \right),$$ The energy required to excite an electron in an atom of hydrogen from $$n = 1$$  to $$n=2$$   is $${\Delta E}$$  (difference in energy $${E_1}$$ and $${E_2}$$
Values of $${E_2}$$ and $${E_1}$$ are,
$$\eqalign{ & {E_1} = - 1.312 \times {10^6}J\,mo{l^{ - 1}} \cr & {E_2} = \frac{{ - 1.312 \times {{10}^6} \times {{\left( 1 \right)}^2}}}{{{{\left( 2 \right)}^2}}} = - 3.28 \times {10^5}J\,mo{l^{ - 1}} \cr & \Delta E\,{\text{is}}\,{\text{given}}\,{\text{by}}\,{\text{the}}\,{\text{relation,}} \cr & \therefore \,\,\Delta E = {E_2} - {E_1} = \left[ { - 3.28 \times {{10}^5}} \right] - \left[ { - 1.312 \times {{10}^6}} \right]\,J\,mo{l^{ - 1}} \cr & = \left( { - 3.28 \times {{10}^5} + 1.312 \times {{10}^6}} \right)\,J\,mo{l^{ - 1}} \cr & = 9.84 \times {10^5}\,J\,mo{l^{ - 1}} \cr} $$
Thus the correct answer is (D)

In $${d_{{z^2}}}$$  orbital two lobes are lying along $$z$$ - axis and it contains a ring of negative charge surrounding the nucleus in $$xy$$ - plane.