1.
A neutron travelling with a velocity $$v$$ and kinetic energy $$E$$ has a perfectly elastic head-on collision with a nucleus of an atom of mass number $$A$$ at rest. The fraction of total energy retained by the neutron is approximately
3.
An $$\alpha $$ particle passes rapidly through the exact centre of a hydrogen molecule, moving on a line perpendicular to the inter-nuclear axis. The distance between the nuclei is $$b.$$ Where on its path does the $$\alpha $$ particle experience the greatest force? (Assume that the nuclei do not move much during the passage of the $$\alpha $$ particle. Also neglect the electric field of the electrons in the molecule.)
Speed of electron in first orbit $$\left( {n = 1} \right)$$ of hydrogen atom $$\left( {z = 1} \right),$$
$$\eqalign{
& v = \frac{{{e^2}}}{{2{\varepsilon _0}h}} \cr
& r = \frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}} \Rightarrow {\varepsilon _0} = \frac{{r\pi m{e^2}}}{{{h^2}}} \cr} $$
Acceleration of electron,
$$\eqalign{
& \frac{{{v^2}}}{r} = \frac{{{e^4}}}{{4\varepsilon _0^2{h^2}}} \times \frac{{\pi m{e^2}}}{{{h^2}{\varepsilon _0}}} \cr
& = \frac{{{e^4} \times \pi m{e^2}}}{{4{h^4}\varepsilon _0^2}}\,......\left( {{\text{ii}}} \right) \cr} $$
eliminating $${\varepsilon _0}$$
$$ = \frac{{{e^4}\pi m{e^2}{h^6}}}{{4{h^4}{r^3}{\pi ^3}{m^3}{e^6}}} = \frac{{{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}$$
5.
Taking the wavelength of first Balmer line in hydrogen
spectrum ($$n = 3$$ to $$n = 2$$ ) as $$660\,nm,$$ the wavelength of the $${2^{nd}}$$ Balmer line ($$n = 4$$ to $$n = 2$$ ) will be;
6.
Electrons are bombarded to excite hydrogen atoms and six spectral lines are observed. If $${E_g}$$ is the ground state energy of hydrogen, the minimum energy the bombarding electrons should posses is
To obtain 6 spectral line, as electron must be excited to fourth orbit with energy $$\frac{{{E_g}}}{{16}},$$ so that the difference is $$\frac{{15\,{E_g}}}{{16}}.$$
7.
The ionization energy of a hydrogen-like Bohr atom is $$4$$ Rydbergs. Find the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state:
$$[1\,Rydberg = 2.2 \times {10^{ - 18}},h = 6.6 \times {10^{ - 34}}Js,c = 3 \times {10^8}m/s.$$ Bohr radius of hydrogen atom $$ = 5 \times {10^{ - 11}}m$$ ]
The energy in the ground state
$$\eqalign{
& {E_1} = - 4\,Rydberg \cr
& {E_1} = - 4 \times 2.2 \times {10^{ - 18}}J \cr} $$
The energy of the first excited state $$\left( {n = 2} \right)$$
$${E_2} = \frac{{{E_1}}}{4} = - 2.2 \times {10^{ - 18}}J$$
The energy difference
$$\Delta E = {E_2} - {E_1} = 3 \times 2.2 \times {10^{ - 18}}J$$
Now, the wavelength of radiation emitted is
$$\eqalign{
& \lambda = \frac{{hc}}{{\Delta E}} \cr
& \lambda = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3 \times 2.2 \times {{10}^{ - 18}}}} = 300\,\mathop {\text{A}}\limits^ \circ \cr} $$
8.
The energy difference between the first two levels of hydrogen atom is $$10.2\,eV$$ for another element of atomic number 10 and mass number 20, this will be
9.
In a Rutherford scattering experiment when a projectile of charge $${Z_1}$$ and mass $${M_1}$$ approaches a target nucleus of charge $${Z_2}$$ and mass $${M_2},$$ the distance of closest approach is $${r_0}.$$ The energy of the projectile is
A.
directly proportional to $${M_1} \times {M_2}$$
A particle of mass $${M_1}$$ and charge $${Z_1}$$ possess initial velocity $$u,$$ when it is at a large distance from the nucleus of an atom having atomic number $${Z_2}.$$ At the distance of closest approach, the kinetic energy of particle is completely converted to potential energy. Mathematically,
$$\frac{1}{2}{M_1}{u^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Z_1}{Z_2}}}{{{r_0}}}$$
So, the energy of the particle is directly proportional to $${Z_1}{Z_2}.$$
10.
An energy of $$24.6\,eV$$ is required to remove one of the electrons from a neutral helium atom. The energy in $$\left( {eV} \right)$$ required to remove both the electrons from a neutral helium atom is
When one $${e^ - }$$ is removed from neutral helium atom, it becomes a one $${e^ - }$$ species.
For one $${e^ - }$$ species we know
$${E_n} = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV/atom$$
For helium ion, $$Z = 2$$ and for first orbit $$n = 1.$$
$$\therefore {E_1} = \frac{{ - 13.6}}{{{{\left( 1 \right)}^2}}} \times {2^2} = - 54.4eV$$
$$\therefore $$ Energy required to remove this $${e^ - } = + 54.4\,eV$$
$$\therefore $$ Total energy required $$ = 54.4 + 24.6 = 79eV$$