Chapter 12: Atoms

Ans: The sizes of the atoms in Thomson's model are no different from the atomic size in Rutherford's model.

Ans: In the ground state of Thomson's model, the electrons are in stable equilibrium, while in Rutherford's model, electrons always experience a net force.

Ans: A classical atom based on Rutherford's model is doomed to collapse.

Ans: An atom features a nearly continuous mass distribution in Thomson's model, but features a highly non-uniform mass distribution in Rutherford's model.

e. The positively charged part of the atom possesses most of the mass in _______ (Rutherford's model/ Thomson's model /both the models.)

Ans: The positively charged part of the atom possesses most of the mass in both the models.

Ans: In the $\alpha -$particle scattering experiment, when a thin sheet of solid hydrogen is replaced with the gold foil, the scattering angle would not turn out to be large enough.

This is because the mass of hydrogen is smaller than the mass of incident $\alpha -$particles. Also, the mass of the scattering particle is more than the target nucleus (hydrogen).

As a consequence, the $\alpha -$particles would not bounce back when solid hydrogen is utilized in the $\alpha -$particle scattering experiment and hence, we cannot evaluate the size of the hydrogen nucleus.

Ans: We know the Rydberg’s formula is given as;

$\frac{hc}{\lambda }=21.76\times {{10}^{-19}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]$

Here,

$h$ = Planck’s constant = $6.6\times {{10}^{-34}}Js$

$c$ = speed of light = $3\times {{10}^{8}}m/s$

(${{n}_{1}}$ and ${{n}_{2}}$ are integers)

The shortest wavelength present within the Paschen series of the spectral lines is for values ${{n}_{1}}=3$ and ${{n}_{2}}=\infty $.

$\Rightarrow$ $\frac{hc}{\lambda }=21.76\times {{10}^{-19}}\left[ \frac{1}{{3}^{2}}-\frac{1}{{\infty}^{2}} \right]$

$\Rightarrow$ $\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}\times 9}{21.76\times {{10}^{-19}}}$

$\Rightarrow$ $\lambda =8.189\times {{10}^{-7}}m$

$\Rightarrow$ $\lambda =818.9nm$

Ans: Given that the distance between the two energy levels in an atom is $E=2.3eV$.

$\Rightarrow E=2.3\times 1.6\times {{10}^{-19}}$

$\Rightarrow E=3.68\times {{10}^{-19}}J$

Let $\nu $ be the frequency of radiation emitted when the atom jumps from the upper level to the lower level.

The relation for energy is given as;

$E=h\nu $

Here,

$h=$ Planck’s constant $=6.6\times {{10}^{-34}}Js$

$\Rightarrow \nu =\frac{E}{h}$

$\Rightarrow \nu =\frac{3.38\times {{10}^{-19}}}{6.62\times {{10}^{-32}}}$

$\Rightarrow \nu =5.55\times {{10}^{14}}Hz$

Clearly, the frequency of the radiation is $5.6\times {{10}^{14}}Hz$.

Ans: Provided that the ground state energy of hydrogen atom, $E=-13.6eV$which is the total energy of a hydrogen atom. 

Here, kinetic energy is equal to the negative of the total energy.

Kinetic energy $=-E=-\left( -13.6 \right)=13.6eV$

The potential energy is the same as the negative of two times kinetic energy.

Potential energy $=-2\times \left( 13.6 \right)=-27.2eV$

$\therefore $ The kinetic energy of the electron is $13.6eV$ and the potential energy is $-27.2eV$ .

Ans: It is known that for ground level absorption, ${{n}_{1}}=1$

Let ${{E}_{1}}$ be the energy of this level. It is known that ${{E}_{1}}$ is related with ${{n}_{1}}$ as;

${{E}_{1}}=\frac{-13.6}{n_{1}^{2}}eV$

$\Rightarrow {{E}_{1}}=\frac{-13.6}{{{1}^{2}}}=-13.6eV$

When the atom jumps to a higher level, ${{n}_{2}}=4$.

Let ${{E}_{2}}$ be the energy of this level.

$\Rightarrow {{E}_{2}}=\frac{-13.6}{n_{2}^{2}}eV$

$\Rightarrow {{E}_{2}}=\frac{-13.6}{{{4}^{2}}}=\frac{-13.6}{16}eV$

The amount of energy absorbed by the photon is given as;

\[E={{E}_{1}}-{{E}_{2}}\]

\[\Rightarrow E=\left( \frac{-13.6}{16} \right)-\left( \frac{-13.6}{1} \right)\]

\[\Rightarrow E=\frac{13.6\times 15}{16}eV\]

\[\Rightarrow E=\frac{13.6\times 16}{16}\times 1.6\times {{10}^{-19}}\]

\[\Rightarrow E=2.04\times {{10}^{-18}}J\]

For a photon of wavelength $\lambda $ , the expression of energy is written as;

$E=\frac{hc}{\lambda }$

Here,

$h=$ Planck’s constant $=6.6\times {{10}^{-34}}Js$

$c=$ speed of light $=3\times {{10}^{8}}m/s$

$\Rightarrow \lambda =\frac{hc}{E}$

$\Rightarrow \lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.04\times {{10}^{-18}}}$

$\Rightarrow \lambda =9.7\times {{10}^{-8}}m$

$\Rightarrow \lambda =97nm$

Also, frequency of a photon is given by the relation,

$\nu =\frac{c}{\lambda }$

$\Rightarrow \nu =\frac{3\times {{10}^{8}}}{9.7\times {{10}^{-8}}}\approx 3.1\times {{10}^{15}}Hz$

Clearly, the wavelength of the photon is 97nm whereas the frequency is $3.1\times {{10}^{15}}Hz$.

Ans: Consider ${{\nu }_{1}}$ to be the orbital speed of the electron in a hydrogen atom in the ground state level ${{n}_{1}}=1$ . For charge $\left( e \right)$ of an electron, ${{\nu }_{1}}$ is given by the relation,

${{\nu }_{1}}=\frac{{{e}^{2}}}{{{n}_{1}}4\pi {{\in }_{0}}\left( \frac{h}{2\pi } \right)}$

$\Rightarrow {{\nu }_{1}}=\frac{{{e}^{2}}}{2{{\in }_{0}}h}$

Here,

$e=1.6\times {{10}^{-19}}C$

${{\in }_{0}}=$Permittivity of free space $=8.85\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

$h=$ Planck’s constant $=6.6\times {{10}^{-34}}Js$

$\Rightarrow {{\nu }_{1}}=\frac{{{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{2\times 8.85\times {{10}^{-12}}\times 6.62\times {{10}^{-34}}}$

$\Rightarrow {{\nu }_{1}}=0.0218\times {{10}^{8}}$

$\Rightarrow {{\nu }_{1}}=2.18\times {{10}^{6}}m/s$

For level ${{n}_{2}}=2$ , we can write the relation for the corresponding orbital speed as;

${{\nu }_{2}}=\frac{{{e}^{2}}}{{{n}_{2}}2{{\in }_{0}}h}$

$\Rightarrow {{\nu }_{2}}=\frac{{{\left( 1.16\times {{10}^{-19}} \right)}^{2}}}{2\times 2\times 8.85\times {{10}^{-12}}\times 6.62\times {{10}^{-34}}}$

$\Rightarrow {{\nu }_{2}}=1.09\times {{10}^{6}}m/s$

And, for ${{n}_{3}}=3$ , we can write the relation for the corresponding orbital speed as;

${{\nu }_{3}}=\frac{{{e}^{2}}}{{{n}_{3}}2{{\in }_{0}}h}$

$\Rightarrow {{\nu }_{3}}=\frac{{{\left( 1.16\times {{10}^{-19}} \right)}^{2}}}{3\times 2\times 8.85\times {{10}^{-12}}\times 6.62\times {{10}^{-34}}}$

$\Rightarrow {{\nu }_{3}}=7.27\times {{10}^{5}}m/s$

Clearly, the speeds of the electron in a hydrogen atom in the levels $n=1,2$ and $3$ are $2.18\times {{10}^{6}}m/s$ , $1.09\times {{10}^{6}}m/s$ and $7.27\times {{10}^{5}}m/s$ respectively.

Ans: Consider ${{T}_{1}}$ to be the orbital period of the electron when it is in level ${{n}_{1}}=1$ .

It is known that the orbital period is related to the orbital speed as

${{T}_{1}}=\frac{2\pi {{r}_{1}}}{{{\nu }_{1}}}$

Here,

${{r}_{1}}=$ Radius of the orbit in ${{n}_{1}}$$=\frac{n_{1}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}$

$h=$ Planck’s constant $=6.6\times {{10}^{-34}}Js$

$e=$ Charge of an electron $=1.6\times {{10}^{-19}}C$

${{\in }_{0}}=$ Permittivity of free space $=8.85\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

$m=$ Mass of an electron $=9.1\times {{10}^{-31}}kg$

$\Rightarrow {{T}_{1}}=\frac{2\pi \times {{\left( 1 \right)}^{2}}\times {{\left( 6.62\times {{10}^{-34}} \right)}^{2}}\times 8.85\times {{10}^{-12}}}{2.18\times {{10}^{6}}\times \pi \times 9.1\times {{10}^{-31}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}$

$\Rightarrow {{T}_{1}}=15.27\times {{10}^{-17}}$

$\Rightarrow {{T}_{1}}=1.527\times {{10}^{-16}}s$

For level ${{n}_{2}}=2$, we can write the orbital period as;

${{T}_{2}}=\frac{2\pi {{r}_{2}}}{{{\nu }_{2}}}$

Here,

${{r}_{2}}=$ Radius of the orbit in ${{n}_{2}}$$=\frac{n_{2}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}$

$\Rightarrow {{T}_{1}}=\frac{2\pi \times {{\left( 2 \right)}^{2}}\times {{\left( 6.62\times {{10}^{-34}} \right)}^{2}}\times 8.85\times {{10}^{-12}}}{1.09\times {{10}^{6}}\times \pi \times 9.1\times {{10}^{-31}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}=1.22\times {{10}^{-15}}s$

And for the level ${{n}_{3}}=3$, we can write the orbital period as;

${{T}_{3}}=\frac{2\pi {{r}_{3}}}{{{\nu }_{3}}}$

Here,

${{r}_{3}}=$ Radius of the orbit in ${{n}_{3}}$$=\frac{n_{3}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}$

$\Rightarrow {{T}_{3}}=\frac{2\pi \times {{\left( 3 \right)}^{2}}\times {{\left( 6.62\times {{10}^{-34}} \right)}^{2}}\times 8.85\times {{10}^{-12}}}{7.27\times {{10}^{5}}\times \pi \times 9.1\times {{10}^{-31}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}=4.12\times {{10}^{-15}}s$

Hence, the orbital periods in the levels $n=1,2$ and $3$ are $1.527\times {{10}^{-16}}s$ , $1.22\times {{10}^{-15}}s$ and $4.12\times {{10}^{-15}}s$ respectively.

Ans: Provided that the innermost radius, ${{r}_{1}}=5.3\times {{10}^{-11}}m$ .

Let ${{r}_{2}}$ be the radius of the orbit at $n=2$ . It is related to the radius of the innermost orbit as;

${{r}_{2}}={{\left( n \right)}^{2}}{{r}_{1}}$

$\Rightarrow {{r}_{2}}={{\left( 2 \right)}^{2}}\times 5.3\times {{10}^{-11}}=2.1\times {{10}^{-10}}m$

Similarly, for $n=3$;

${{r}_{3}}={{\left( n \right)}^{2}}{{r}_{1}}$

$\Rightarrow {{r}_{3}}={{\left( 3 \right)}^{2}}\times 5.3\times {{10}^{-11}}=4.77\times {{10}^{-10}}m$

Clearly, the radii of the $n=2$ and $n=3$ orbits are $2.1\times {{10}^{-10}}m$ and $4.77\times {{10}^{-10}}m$ respectively.

Ans: It is provided that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is $12.5eV$ .

It is also known that the energy of the gaseous hydrogen in its ground state at room temperature is $-13.6eV$ .

When gaseous hydrogen is bombarded with an electron beam at room temperature, the energy of the gaseous hydrogen becomes \[-13.6+12.5eV=-1.1eV\].

Now, the orbital energy is related to orbit level \[\left( n \right)\] as;

$E=\frac{-13.6}{{{\left( n \right)}^{2}}}eV$

For $n=3;E=\frac{-13.6}{9}=-1.5eV$

This energy is approximately equal to the energy of gaseous hydrogen.

So, it can be concluded that the electron has excited from \[n=1\text{ }to\text{ }n=3\] level.

During its de-excitation, the electrons can jump from \[n=3\text{ }to\text{ }n=1\] directly, which forms a line of the Lyman series of the hydrogen spectrum.

The formula for wave number for Lyman series is given as;

$\frac{1}{\lambda }={{R}_{y}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right)$

Here, 

${{R}_{y}}=$ Rydberg constant $=1.097\times {{10}^{7}}{{m}^{-1}}$

\[\lambda =\] Wavelength of radiation emitted by the transition of the electron

Using this relation for $n=3$ we get,

$\frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right)$

\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( 1-\frac{1}{9} \right)\]

\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{8}{9} \right)\]

$\Rightarrow \lambda =\frac{9}{8\times 1.097\times {{10}^{7}}}=102.55nm$

If the transition takes place from $n=3\text{ }to\text{ }n=2$ , and then from \[n=2\text{ }to\text{ }n=1\], then the wavelength of the radiation emitted in transition from \[n=3\text{ }to\text{ }n=2\] is given as;

$\frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)$

\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{4}-\frac{1}{9} \right)\]

\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{5}{36} \right)\]

$\Rightarrow \lambda =\frac{36}{5\times 1.097\times {{10}^{7}}}=656.33nm$

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Now, the wavelength of the radiation when the transition takes place from \[n=2\text{ }to\text{ }n=1\] is given as;

$\frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)$

\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( 1-\frac{1}{4} \right)\]

\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{3}{4} \right)\]

$\Rightarrow \lambda =\frac{4}{3\times 1.097\times {{10}^{7}}}=121.54nm$

Clearly, in the Lyman series, two wavelengths are emitted i.e., 102.5nm and 121.5nm whereas in the Balmer series, only one wavelength is emitted i.e., 656.33nm.

Ans: Here, it is provided that,

Radius of the earth’s orbit around the sun, $r=1.5\times {{10}^{11}}m$

Orbital speed of the earth, $\nu =3\times {{10}^{4}}m/s$

Mass of the earth, $m=6\times {{10}^{24}}kg$

With respect to the Bohr’s model, angular momentum is quantized and is given as;

$m\nu r=\frac{nh}{2\pi }$

Here,

$h=$ Planck’s constant $=6.6\times {{10}^{-34}}Js$

$n=$ Quantum number

$n=\frac{m\nu {{r}^{2}}\pi }{h}$

$\Rightarrow n=\frac{2\pi \times 6\times {{10}^{24}}\times 3\times {{10}^{4}}\times 1.5\times {{10}^{11}}}{6.62\times {{10}^{-34}}}$

$\Rightarrow n=25.61\times {{10}^{73}}=2.6\times {{10}^{74}}$

Clearly, the quantum number that characterizes the earth’s revolution around the sun is $2.6\times {{10}^{74}}$ .