Atoms And Nuclei MCQ Questions & Answers in Modern Physics | Physics

In hydrogen atom, wavelength of characteristic spectrum
$$\frac{1}{\lambda } = R{z^2}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$
For Lyman series $${n_1} = 1,{n_2} = 2$$
$$\frac{1}{{{\lambda _1}}} = R{z^2}\left[ {\frac{1}{{{{\left( 1 \right)}^2}}} - \frac{1}{{{{\left( 2 \right)}^2}}}} \right]\,......\left( {\text{i}} \right)$$
For Balmer series $${n_1} = 2,{n_2} = 3$$
$$\frac{1}{{{\lambda _2}}} = R{z^2}\left[ {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 3 \right)}^2}}}} \right]\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i) we get
$$\eqalign{ & \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{R{z^2}\left[ {\frac{1}{4} - \frac{1}{9}} \right]}}{{R{z^2}\left[ {1 - \frac{1}{4}} \right]}} = \frac{{\frac{5}{{36}}}}{{\frac{3}{4}}} \cr & \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{5}{{36}} \times \frac{4}{3} = \frac{5}{{27}} \cr} $$

Nuclear density of an atom of mass number $$A,$$
$$\eqalign{ & d = \frac{{{\text{ mass }}}}{{{\text{ volume }}}} = \frac{{A\left( {1.67 \times {{10}^{ - 27}}} \right)}}{{\frac{4}{3}\pi {{\left[ {1.25 \times {{10}^{ - 15}}{A^{\frac{1}{3}}}} \right]}^3}}} \cr & \left[ {\because V = \frac{4}{3}\pi {R^3},R = {R_0}{A^{\frac{1}{3}}},{R_0} = 1.25 \times {{10}^{ - 15}}} \right] \cr & \therefore d = 2 \times {10^{17}}kg/{m^3} \cr} $$

Excitation energy $$\Delta E = {E_2} - {E_1} = 13.6\,{Z^2}\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right]$$
$$\eqalign{ & \Rightarrow 40.8 = 13.6 \times \frac{3}{4} \times {Z^2} \cr & \Rightarrow Z = 2. \cr} $$
Now required energy to remove the electron from ground state
$$ = \frac{{ + 13.6{Z^2}}}{{{{\left( 1 \right)}^2}}} = 13.6{\left( Z \right)^2} = 54.4\,eV.$$

KEY CONCEPT : The maximum number of electrons in an orbit is $$2{n^2}.\,n > 4$$   is not allowed.
Therefore the number of maximum electron that can be in first four orbits are
$$\eqalign{ & 2{\left( 1 \right)^2} + 2{\left( 2 \right)^2} + 2{\left( 3 \right)^2} + 2{\left( 4 \right)^2} \cr & = 2 + 8 + 18 + 32 = 60 \cr} $$
Therefore, possible element are 60.

Magnetic field at the centre of nucleus of $$H$$-atom,
$$B = \frac{{{\mu _0}I}}{{2r}}\,......\left( {\text{i}} \right)$$
According to Bohr's model, radius of orbit $$r \propto {n^2}$$  from eq. (i) we can also write as $$B \propto {n^{ - 2}}$$

Number of possible spectral lines emitted when an electron jumps back to ground state from n^th orbit $$ = \frac{{n\left( {n - 1} \right)}}{2}$$
Here, $$\frac{{n\left( {n - 1} \right)}}{2} = 6$$
$$ \Rightarrow n = 4$$
Wavelength $$\lambda $$ from transition from $$n= 1$$  to $$n= 4$$  is given by,
$$\eqalign{ & \frac{1}{\lambda } = R\left( {\frac{1}{1} - \frac{1}{{{4^2}}}} \right) \cr & \Rightarrow \lambda = \frac{{16}}{{15R}} = 975\,\mathop {\text{A}}\limits^ \circ \cr} $$

$$\eqalign{ & {\text{Using}}\,\frac{1}{\lambda } = R{\left( {Z - 1} \right)^2}\left[ {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right] \cr & {\text{For}}\,\alpha \,{\text{particle,}}\,{n_1} = 2,{n_2} = 1 \cr} $$
For metal $$A$$ :
$$\frac{{1875\,R}}{4} = R{\left( {{Z_1} - 1} \right)^2}\left( {\frac{3}{4}} \right) \Rightarrow {Z_1} = 26$$
For metal $$B$$ :
$$675\,R = R{\left( {{Z_2} - 1} \right)^2}\left( {\frac{3}{4}} \right) \Rightarrow {Z_2} = 32$$
Therefore, 4 elements lie between $$A$$ and $$B.$$

According to third postulate of Bohr’s model, when an atom makes a transition from higher energy state to lower energy state, the difference of energy is carried away by a photon such that $$h\nu = {E_{{n_i}}} - {E_{{n_f}}}\,\,{\text{or}}\,\,\frac{{hc}}{\lambda } = Rhc\left[ {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right]$$
[where, $${n_i}$$ = quantum number of higher energy state and $${n_f}$$ = quantum number of lower energy state.]
Energy difference between fifth and first orbit is
$$\eqalign{ & {E_5} - {E_1} = \frac{{hc}}{\lambda }\,\,{\text{and}}\,\,Rhc - \frac{{Rhc}}{{25}} = \frac{{hc}}{\lambda } \cr & \frac{{24}}{{25}}R = \frac{1}{\lambda } \cr & {\text{As,}}\,\,p = \frac{h}{\lambda }\,{\text{and}}\,\nu = \frac{h}{{m\lambda }} \cr & = \frac{{24}}{{25}}\frac{{Rh}}{m} \cr} $$

Energy of n^th state in hydrogen is same as energy of 3n^th state in $$L{i^{ + + }}.$$
$$\therefore 3 \to 1$$   transition in $$H$$ would give same energy as the $$3 \times 3 \to 1 \times 3$$    transitor in $$L{i^{ + + }}.$$

$$\frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$
For limiting wavelength of Lyman series
$$\eqalign{ & {n_1} = 1,{n_2} = \infty \cr & \frac{1}{{{\lambda _L}}} = R \cr} $$
For limiting wavelength of Balmer series
$$\eqalign{ & {n_1} = 2,{n_2} = \infty \cr & \frac{1}{{{\lambda _B}}} = R\left( {\frac{1}{4}} \right) \cr & \Rightarrow {\lambda _B} = \frac{4}{R} \cr & \therefore {\lambda _B} = 4{\lambda _L} = 4 \times 912\,\mathop {\text{A}}\limits^ \circ \cr} .$$