Atomic Structure MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & r \propto {n^2}/Z \cr & {\text{where, }}n = {\text{number of orbit}} \cr & Z = {\text{atomic number}} \cr & \because \,\,{{\text{r}}_1} \propto n_1^2 \cr & {r_2} \propto n_2^2\,\,\left( {Z = 1{\text{ for }}H{\text{ - }}atom} \right) \cr & {\text{So,}}\,\,\,\frac{{{r_1}}}{{{r_2}}} = \frac{{n_1^2}}{{n_2^2}}\,\, \cr & \frac{{0.530}}{{{r_2}}} = \frac{{{1^2}}}{{{2^2}}} \cr & \therefore \,\,\,\,{r_2} = 0.530 \times 4 = 2.120\,\mathop {\text{A}}\limits^{\text{o}} \cr} $$

$$X-$$ rays can ionise gases and cannot get deflected by electric and magnetic fields, wavelength of these rays is 150 to $$0.1\mathop {\text{A}}\limits^{\text{o}} $$ . Thus the wavelength of $$X-$$ rays is shorter than that of $$u.v.$$  rays.

Orbital angular momentum $$ = \sqrt {l\left( {l + 1} \right)} \frac{h}{{2\pi }}$$
Thus, it depends on $$'l'$$  only.

When $$n$$ = 4; $$l$$ = 0, 1, 2, 3
Number of orbitals = 1 + 3 + 5 + 7 = 16
Number of electrons = 2 + 6 + 10 + 14 = 32

$${\text{Given,}}\,{v_A} = 0.1\,m{s^{ - 1}}\,{\text{and}}$$      $${v_B} = 0.05\,m{s^{ - 1}}{\text{also,}}$$     $${m_B} = 5{m_A}$$
$$\eqalign{ & {\text{de - Broglie wavelength,}}\,\lambda = \frac{h}{{mv}} \cr & \therefore \,\,\frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{{\frac{h}{{{m_A}{v_A}}}}}{{\frac{h}{{{m_B}{v_B}}}}} = \frac{{{m_B}{v_B}}}{{{m_A}{v_A}}} \cr & = \frac{{5{m_A} \times 0.05}}{{{m_A} \times 0.1}} = 5 \times 0.5 = 2.5 = \frac{5}{2} \cr & \therefore \,\,{\lambda _A}:{\lambda _B} = 5:2 \cr} $$

$$\eqalign{ & \frac{1}{2}m{\nu ^2} = h\nu - h{\nu _0} \cr & \Rightarrow \frac{1}{2}m{\nu ^2} = h\left( {\nu - {\nu _0}} \right) \cr & \Rightarrow \nu = \sqrt {\frac{{2h}}{m}\left( {\nu - {\nu _0}} \right)} \cr & = \sqrt {\frac{{2 \times 6.6 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}}}}\left( {2 \times {{10}^{15}} - 7 \times {{10}^{14}}} \right)} \cr & = 1.37 \times {10^6}m{{\bar s}^1} \cr} $$

$${N^{3 - }},{F^ - }$$ and $$N{a^ + }$$ contain 10 electrons each.

2^nd excited state will be the 3^rd energy level.
$${E_n} = \frac{{13.6}}{{{n^2}}}eV\,{\text{or}}\,E = \frac{{13.6}}{9}eV = 1.51\,eV$$

TIPS/Formulae : The two guiding rules to arrange the various orbitals in the increasing energy are:
(i) Energy of an orbital increases with increase in the value of $$n + l.$$
(ii) Of orbitals having the same value of $$n + l,$$ the orbital with lower value of n has lower energy.
Thus for the given orbitals, we have
$$\eqalign{ & ({\text{i}})\,\,n + l = 4 + 1 = 5 \cr & {\text{(ii)}}\,n + l = 4 + 0 = 4 \cr & {\text{(iii)}}\,n + l = 3 + 2 = 5 \cr & {\text{(iv)}}\,n + l = 3 + 1 = 4 \cr} $$
Hence, the order of increasing energy is (iv) < (ii) < (iii) < (i)

Bohr's atomic theory is applicable to hydrogen and Single electron species like $$H{e^ + },L{i^{2 + }},B{e^{3 + }},$$    etc.