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

Note that $$v$$ is inversely proportional to $$n$$ whereas $$r$$ is directly to $${n^2}.$$

Ionisation energy is defined as the energy required to knock an electron completely out of an isolated gaseous atom. When electron is raised to the orbit $$n = \infty ,$$  it will be completely out of the atom.
lonisation energy of helium, $$E = \frac{{2{\pi ^2}m{Z^2}{k^2}{e^4}}}{{{h^2}}}\left[ {\frac{1}{{{1^2}}} - \frac{1}{\infty }} \right]$$
$$ = \frac{{2{\pi ^2}m{{\left( 2 \right)}^2}{k^2}{e^4}}}{{{h^2}}}\left[ {\frac{1}{{{1^2}}} - \frac{1}{\infty }} \right]$$
lonisation energy for hydrogen atom
$$\eqalign{ & = 13.6\,eV = \frac{{2{\pi ^2}m{k^2}{e^2}}}{{{h^2}}} \times 4 \times \left[ {\frac{1}{{{1^2}}} - \frac{1}{{{\infty ^2}}}} \right] \cr & \Rightarrow = 4 \times 13.6 = 54.4\,eV \cr} $$

$${E_n} = - 3.4\,eV$$
The kinetic energy is equal to the magnitude of total energy in this case.
$$\therefore K.E. = + 3.4\,eV$$
The de Broglie wavelength of electron
$$\eqalign{ & \lambda = \frac{h}{{\sqrt {2mK} }} = \frac{{6.64 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 3.4 \times 1.6 \times {{10}^{ - 19}}} }} \cr & = 0.66 \times {10^{ - 9}}m \cr} $$

$${E_n} = - \frac{{I.E.}}{{{n^2}}}$$   for Bohr’s hydrogen atom.
Here, $$I.E. = 4R$$
$$\eqalign{ & \therefore {E_n} = \frac{{ - 4R}}{{{n^2}}} \cr & \therefore {E_2} - {E_1} = \frac{{ - 4{R^2}}}{{{2^2}}} - \left( { - \frac{{4{R^2}}}{{{1^2}}}} \right) = 3R\,......\left( {\text{i}} \right) \cr & {E_2} - {E_1} = hv = \frac{{hc}}{\lambda }\,......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii)
$$\eqalign{ & \frac{{hc}}{\lambda } = 3R \cr & \therefore \lambda = \frac{{hc}}{{3R}} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2.2 \times {{10}^{ - 18}} \times 3}} = 300\,\mathop {\text{A}}\limits^ \circ \cr} $$

For transition from $$n = 5$$  to $$n = 4,$$
$$hv = 13.6 \times 9\left[ {\frac{1}{{16}} - \frac{1}{{25}}} \right] = \frac{{13.6 \times 9 \times 9}}{{16 \times 25}} = 2.754\,eV$$
For transition from $$n= 4$$  to $$n=3,$$
$$hv' = 13.6 \times 9\left[ {\frac{1}{9} - \frac{1}{{16}}} \right] = \frac{{13.6 \times 9 \times 7}}{{9 \times 16}} = 5.954\,eV$$
For transition $$n= 4$$  to $$n= 3,$$  the frequency is high and hence wavelength is short.
For photoelectric effect, $$hv' - W = e{V_0},$$
where $$W=$$  work function
$$\eqalign{ & 5.95 \times 1.6 \times {10^{ - 19}} - W = 1.6 \times {10^{ - 19}} \times 3.95 \cr & \Rightarrow W = 2 \times 1.6 \times {10^{ - 19}} = 2\,eV \cr} $$
Again applying $$hv - W = e{{V'}_0}$$
We get, $$2.574 \times 1.6 \times {10^{ - 19}} - 2 \times 1.6 \times {10^{ - 19}} = 1.6 \times {10^{ - 19}}{{V'}_0}$$
$$ \Rightarrow {{V'}_0} = 0.754\,V$$

$$\eqalign{ & \frac{{hc}}{{{\lambda _1}}} - \frac{{hc}}{{{\lambda _0}}} = K.E{._1}\,\,{\text{and}}\,\,\frac{{hc}}{{{\lambda _2}}} - \frac{{hc}}{{{\lambda _0}}} = K.E{._2} \cr & \Rightarrow \frac{{hc}}{{{\lambda _1}}} - \frac{{hc}}{{{\lambda _2}}} = K.E{._1} - K.E{._2} \cr & \Rightarrow hc\left[ {\frac{{{\lambda _2} - {\lambda _1}}}{{{\lambda _1}{\lambda _2}}}} \right] = K.E{._1} - K.E{._2} \cr & \therefore h = \frac{{\left( {K.E{._1} - K.E{._2}} \right){\lambda _1}{\lambda _2}}}{{c\left( {{\lambda _2} - {\lambda _1}} \right)}} \cr} $$

Ionisation energy corresponding to ionisation potential $$\left( {{E_1} = - 13.6\,eV} \right)$$
Photon energy incident $$\left( {\Delta E} \right) = 12.1\,eV$$
So, the energy of electron in excited state $$\left( {{E_2}} \right)$$  is given by
$$\eqalign{ & {E_2} - {E_1} = \Delta E \cr & \Rightarrow {E_2} = \Delta E + {E_1} \cr & \Rightarrow {E_2} = - 13.6 + 12.1 \cr & \Rightarrow {E_2} = - 1.5\,eV \cr & {\text{i}}{\text{.e}}{\text{.}}\,{E_2} = - \frac{{13.6}}{{{n^2}}}eV \cr & - 1.5 = \frac{{ - 13.6}}{{{n^2}}} \cr & \Rightarrow {n^2} = \frac{{ - 13.6}}{{ - 1.5}} \approx 9 \cr & \therefore n = 3 \cr} $$
i.e. energy of electron in excited state corresponds to third orbit.
The possible spectral lines is given by
$$\frac{{n\left( {n - 1} \right)}}{2} \Rightarrow \frac{{3\left( {3 - 1} \right)}}{2} \Rightarrow 3$$

Kinetic energy of electron $$KE = \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}}$$
Potential energy of electron $$U = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Z{e^2}}}{r}$$
∴ Total energy $$E = KE + U = \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}} - \frac{{Z{e^2}}}{{4\pi {\varepsilon _0}r}}$$
$$\eqalign{ & {\text{or}}\,\,E = - \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}}\,\,{\text{or}}\,\,E = - KE \cr & {\text{or}}\,\,KE = - E = - \left( { - 3.4} \right) = 3.4\,eV \cr} $$
NOTE
Total kinetic energy of a revolving electron in any given orbit is equal to the negative of total energy of electrons in that orbit i.e. $$KE = - E$$

$$\eqalign{ & \Delta \lambda = {\lambda _1} - {\lambda _2} \cr & \frac{1}{{R\left[ {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 3 \right)}^2}}}} \right]{Z^2}}} = - \frac{1}{{R\left[ {1 - 0} \right]{Z^2}}} \cr} $$
On solving, $$R = \frac{{31}}{5}\frac{1}{{\Delta \lambda .{Z^2}}}$$

Linear momentum $$ \Rightarrow mv\,\alpha \frac{1}{n}$$
angular momentum $$ \Rightarrow mv\,\alpha n$$
$$\therefore $$ product of linear momentum and angular momentum $$\alpha {n^0}.$$