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

KEY CONCEPT :
The energy of nth orbit of hydrogen is given by
$$\eqalign{ & {E_n} = - \frac{{13.6}}{{{n^2}}}eV/atom \cr & {\text{For}}\,n = 2,{E_n} = \frac{{ - 13.6}}{4}{\text{ = }} - 3.4eV \cr} $$
Therefore the energy required to remove electron from $$n = 2$$  is $$+ 3.4\,eV.$$

Distance of closest approach
$$\eqalign{ & {r_0} = \frac{{Ze\left( {2e} \right)}}{{4\pi {\varepsilon _0}\left( {\frac{1}{2}m{v^2}} \right)}} \cr & {\text{Energy,}}\,\,E = 5 \times {10^6} \times 1.6 \times {10^{ - 19}}J \cr & \therefore {r_0} = \frac{{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)}}{{5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} \cr & \Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm. \cr} $$

We know that $$\frac{1}{\lambda } = R{Z^2}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$
The wave length of first spectral line in the Balmer series of hydrogen atom is 6561 $$\mathop {\text{A}}\limits^ \circ .$$ Here $${n_2} = 3$$  and $${n_1} = 2$$
$$\therefore \frac{1}{{6561}} = R = {\left( 1 \right)^2}\left( {\frac{1}{4} - \frac{1}{9}} \right) = \frac{{5R}}{{36}}\,.....\left( {\text{i}} \right)$$
For the second spectral line in the Balmer series of singly ionised helium ion $${n_2} = 4$$  and $${n_1} = 2;Z = 2$$
$$\therefore \frac{1}{\lambda } = R{\left( 2 \right)^2}\left[ {\frac{1}{4} - \frac{1}{{16}}} \right] = \frac{{3R}}{4}\,.....\left( {{\text{ii}}} \right)$$
Diving equation (i) and equation (ii) we get
$$\eqalign{ & \frac{\lambda }{{6561}} = \frac{{5R}}{{36}} \times \frac{4}{{3R}} = \frac{5}{{27}} \cr & \therefore \lambda = 1215A \cr} $$

The outermost electrons in an atom are called valence electrons. In alkali metals (IA group) outermost electron is present in $$s$$-orbital. Alkali metals have one valence electron. e.g. $$Li,\,Na,\,K.$$

Kinetic energy of electron is given by
$$KE = \frac{{kZ{e^2}}}{{2r}}$$
Potential energy of electron is $$U = - \frac{{kZ{e^2}}}{r}$$
When a hydrogen atom is raised from the ground, to an excited state both potential energy and kinetic energy decreases.

In line emission spectrum, every line spectrum consists of a few isolated bright lines, each bright line corresponds to a particular wavelength. It is emitted by atoms in the gaseous state.
e.g. a sodium discharge lamp, a mercury vapour lamp, a neon discharge tube and a helium discharge tube all emit sharp lines of definite wavelengths.

The possible number of the spectral lines is given
$$ = \frac{{n\left( {n - 1} \right)}}{2} = \frac{{4\left( {4 - 1} \right)}}{2} = 6$$

Radius of n^th orbit $${r_n} \propto {n^2}$$   graph between $${r_n}$$ and $$n$$ is a parabola. Also,
$$\frac{{{r_n}}}{{{r_1}}} = {\left( {\frac{n}{1}} \right)^2} \Rightarrow {\log _e}\left( {\frac{{{r_n}}}{{{r_1}}}} \right) = 2{\log _e}\left( n \right)$$
Comparing this equation with $$y= mx +c,$$
Graph between $${\log _e}\left( {\frac{{{r_n}}}{{{r_1}}}} \right)$$   and $${\log _e}\left( n \right)$$  will be a straight line, passing from origin.
Similarly it can be proved that graph between $${\log _e}\left( {\frac{{{f_n}}}{{{f_1}}}} \right)$$   and $${\log _e}n$$  is a straight line. But with negative slops.

$${E_n} = - 13.6\frac{{\left( {{Z^2}} \right)}}{{\left( {{n^2}} \right)}}eV$$
Therefore, ground state energy of doubly ionized lithium atom $$\left( {Z = 3,n = 1} \right)$$    will be
$${E_1} = \left( { - 13.6} \right)\frac{{{{\left( 3 \right)}^2}}}{{{{\left( 1 \right)}^2}}} = - 122.4\,eV$$
$$\therefore $$ Ionization energy of an electron in ground state of doubly ionized lithium atom will be $$122.4\,eV.$$

Wave number $$\frac{1}{\lambda } = R{Z^2}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{{n^2}}}} \right]$$
$$ \Rightarrow \lambda \propto \frac{1}{{{Z^2}}}$$
By question $$n = 1$$  and $${n_1} = 2$$
Then, $${\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}$$