Chapter 12: Atoms

Ans: About the same.

The average angle of deflection of $\alpha -$particles caused by a thin gold foil considered by Thomson's model is about the same size as that considered by Rutherford's model.

This is because in both the models, the average angle was used.

Ans: Much less.

The probability of scattering of $\alpha -$particles at angles greater than 90° considered by Thomson's model is much less than that considered by Rutherford's model.

Ans: Scattering is mainly caused due to single collisions.

The chances of a single collision have a linearly increasing nature with the number of target atoms.

As the number of target atoms increases with an increase in thickness, the collision probability varies linearly with the thickness of the target.

Ans: Thomson's model.

It is incorrect to not consider multiple scattering in Thomson's model for the calculation of average angle of scattering of $\alpha -$particles by a thin foil.

This is because a single collision produces very little deflection in this model.

Thus, the observed average scattering angle can be demonstrated only by considering multiple scattering.

Ans: The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about ${{10}^{-40}}$. 

It is given that,

Radius of the first Bohr orbit is given by the relation,

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

Here,

${{\in }_{0}}=$ Permittivity of free space

$h=$ Planck’s constant $=6.63\times {{10}^{-34}}Js$

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

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

${{m}_{p}}=$ Mass of a proton $=1.67\times {{10}^{-27}}kg$

$r=$ Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given by;

${{F}_{C}}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}{{r}^{2}}}$ …(2)

Gravitational force of attraction between an electron and a proton is given by;

${{F}_{G}}=\frac{G{{m}_{p}}{{m}_{e}}}{{{r}^{2}}}$ …(3)

Here,

$G=$ Gravitational constant $=6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}$

The electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write;

${{F}_{G}}={{F}_{C}}$

$\Rightarrow \frac{G{{m}_{p}}{{m}_{e}}}{{{r}^{2}}}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}{{r}^{2}}}$ …(using 2 & 3)

$\Rightarrow \frac{{{e}^{2}}}{4\pi {{\in }_{0}}}=G{{m}_{p}}{{m}_{e}}$ …(4)

Now, by substituting the value of equation (4) in (1),  we get; 

${{r}_{1}}=\frac{{{\left( \frac{h}{2\pi } \right)}^{2}}}{G{{m}_{p}}{{m}_{e}}}$

$\Rightarrow {{r}_{1}}=\frac{{{\left( \frac{6.63\times {{10}^{-34}}}{2\times 3.14} \right)}^{2}}}{6.67\times {{10}^{-11}}\times 1.67\times {{10}^{-27}}\times {{\left( 9.1\times {{10}^{-31}} \right)}^{2}}}$

$\Rightarrow {{r}_{1}}\approx 1.21\times {{10}^{29}}m$

It is known that the universe is 156 billion light years wide or $1.5\times {{10}^{27}}m$ wide. Clearly, it can be concluded that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Ans: It is given that a hydrogen atom makes transition from an upper level $\left( n \right)$ to a lower level $\left( n-1 \right)$. We have the relation for energy $\left( {{E}_{1}} \right)$ of radiation at level $n$ as;

${{E}_{1}}=h{{\nu }_{1}}=\frac{hm{{e}^{4}}}{{{\left( 4\pi  \right)}^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}}\times \left( \frac{1}{{{n}^{2}}} \right)$ …(1)

Here,

${{\nu }_{1}}=$ Frequency of radiation at level $n$

$h=$ Planck’s constant

$m=$ Mass of hydrogen atom

$e=$ Charge on an electron

${{\in }_{0}}=$ Permittivity of free space

Now, the relation for energy $\left( {{E}_{2}} \right)$ of radiation at level $\left( n-1 \right)$ is given as;

${{E}_{2}}=h{{\nu }_{2}}=\frac{hm{{e}^{4}}}{{{\left( 4\pi  \right)}^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}}\times \frac{1}{{{\left( n-1 \right)}^{2}}}$ …(2)

Here,

${{\nu }_{2}}=$ Frequency of radiation at level $\left( n-1 \right)$

Energy $\left( E \right)$ released as a result of de-excitation;

$E={{E}_{2}}-{{E}_{1}}$

$\Rightarrow h\nu ={{E}_{2}}-{{E}_{1}}$ …(3)

Here,

$\nu =$ Frequency of radiation emitted

Using the values of equations (1) and (2) in equation (3), we get;

$\nu =\frac{m{{e}^{4}}}{{{\left( 4\pi  \right)}^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}}\left[ \frac{1}{{{\left( n-1 \right)}^{2}}}-\frac{1}{{{n}^{2}}} \right]$

$\Rightarrow \nu =\frac{m{{e}^{4}}\left( 2n-1 \right)}{{{\left( 4\pi  \right)}^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}{{n}^{2}}{{\left( n-1 \right)}^{2}}}$

For large $n$, we can write $\left( 2n-1 \right)-2n$ and $\left( n-1 \right)\simeq n$

$\Rightarrow \nu =\frac{m{{e}^{4}}}{32{{\pi }^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}{{n}^{3}}}$ …(4)

Classical relation of frequency of revolution of an electron is given by;

${{\nu }_{c}}=\frac{\nu }{2\pi r}$ …(5)

Here,

Velocity of the electron in the ${{n}^{th}}$ orbit is given as;

$\nu =\frac{{{e}^{2}}}{4\pi {{\in }_{0}}\left( \frac{h}{2\pi } \right)n}$ …(6)

And, radius of the ${{n}^{th}}$ orbit is given by;

$r=\frac{4\pi {{\in }_{0}}{{\left( \frac{h}{2\pi } \right)}^{2}}}{m{{e}^{2}}}{{n}^{2}}$ …(7)

Substituting the values of equation (6) and (7) in equation (5), we get;

${{\nu }_{c}}=\frac{m{{e}^{4}}}{32{{\pi }^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}{{n}^{3}}}$ …(8)

Clearly, when equations (4) and (8) are compared, it can be seen that the frequency of radiation emitted by the hydrogen atom is the same as the classical orbital frequency.

Ans: To construct a quantity with the dimensions of length from the fundamental constants $e,{{m}_{e}}\text{ }and\text{ }c$, take: 

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

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

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

The quantity having dimensions of length and involving the given quantities is $\left( \frac{{{e}^{2}}}{4\pi {{\in }_{0}}{{m}_{e}}{{c}^{2}}} \right)$

Here,

${{\in }_{0}}=$ Permittivity of free space

And, $\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

$\therefore $The numerical value of the taken quantity will be

\[\frac{1}{4\pi {{\in }_{0}}}\times \frac{{{e}^{2}}}{{{m}_{e}}{{c}^{2}}}=9\times {{10}^{9}}\times \frac{{{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{9.1\times {{10}^{-31}}\times {{\left( 3\times {{10}^{8}} \right)}^{2}}}=2.81\times {{10}^{-15}}m\]

Clearly, the numerical value of the taken quantity is much less than the typical size of an atom.

Ans: To construct a quantity with the dimension of length from \[h,{{m}_{e}}\text{ }and\text{ }e\], take,

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

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

Planck’s constant, $h=6.63\times {{10}^{-34}}Js$

Now, let us take a quantity involving the given quantities as, $\frac{4\pi {{\in }_{0}}\left( \frac{{{h}^{2}}}{2\pi } \right)}{{{m}_{e}}{{e}^{2}}}$

Here, ${{\in }_{0}}=$ Permittivity of free space

And, $\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

$\therefore $The numerical value of the taken quantity would be

\[4\pi {{\in }_{0}}\times \frac{{{\left( \frac{h}{2\pi } \right)}^{2}}}{{{m}_{e}}{{e}^{2}}}=\frac{1}{9\times {{10}^{9}}}\times \frac{{{\left( \frac{6.63\times {{10}^{-34}}}{2\times 3.14} \right)}^{2}}}{9.1\times {{10}^{-31}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}=0.53\times {{10}^{-10}}m\]

Clearly, the value of the quantity taken is of the order of the atomic size.

Ans: Kinetic energy of the electron is the same as the negative of the total energy.

$\Rightarrow K=-E$

\[\Rightarrow K=-\left( -3.4 \right)=+3.4eV\]

Clearly, the kinetic energy of the electron in the given state is $+3.4eV$.

Ans: Potential energy $\left( U \right)$ of the electron is the same as the negative of twice of its kinetic energy

$\Rightarrow U=-2K$

\[\Rightarrow U=-\left( -3.4 \right)=-6.8eV\]

Clearly, the potential energy of the electron in the given state is $-6.8eV$.

Ans: The potential energy of a system is dependent on the reference point taken. Here, the potential energy of the reference point is considered to be zero.

When the reference point is changed, then the magnitude of the potential energy of the system also changes.

As total energy is the sum of kinetic and potential energies, total energy of the system would also differ.

Ans: It is not much spoken about the quantization of orbits of planets around the Sun since the angular momentum associated with planetary motion is largely relative to the value of constant $\left( h \right)$. 

The angular momentum of the Earth in its orbit is of the order of ${{10}^{70}}h$. This causes a very high value of quantum levels $n$ of the order of ${{10}^{70}}$. 

When large values of $n$ are considered, successive energies and angular momenta are found to be relatively very small. Clearly, the quantum levels for planetary motion are considered continuous.

Ans: To obtain the first Bohr's radius and the ground state energy of a muonic hydrogen atom, consider the mass of a negatively charged muon to be ${{m}_{\mu }}=207m$.

Now, according to Bohr’s model,

Bohr radius, ${{r}_{e}}\propto \left( \frac{1}{{{m}_{e}}} \right)$

And, energy of a ground state electronic hydrogen atom, ${{E}_{e}}\propto {{m}_{e}}$.

Also, energy of a ground state muonic hydrogen atom, ${{E}_{\mu }}\propto {{m}_{e}}$.

It is known that the value of the first Bohr orbit, ${{r}_{e}}=0.53A=0.53\times {{10}^{-10}}m$

Consider ${{r}_{\mu }}$ to be the radius of muonic hydrogen atom.

At equilibrium, we have the relation:

${{m}_{\mu }}{{r}_{\mu }}={{m}_{e}}{{r}_{e}}$

$\Rightarrow 207{{m}_{e}}\times {{r}_{\mu }}={{m}_{e}}{{r}_{e}}$

$\Rightarrow {{r}_{\mu }}=\frac{0.53\times {{10}^{-10}}}{207}=2.56\times {{10}^{-13}}m$

Clearly, the value of the first Bohr radius of muonic hydrogen atom is $2.56\times {{10}^{-13}}m$.

Now, we have, 

${{E}_{e}}=-13.6eV$
Taking the ratio of these energies as $\frac{{{E}_{e}}}{{{E}_{\mu }}}=\frac{{{m}_{e}}}{{{m}_{\mu }}}$

$\Rightarrow \frac{{{E}_{e}}}{{{E}_{\mu }}}=\frac{{{m}_{e}}}{207{{m}_{e}}}$

$\Rightarrow {{E}_{\mu }}=207{{E}_{e}}$

$\Rightarrow {{E}_{\mu }}=207\times \left( -13.6 \right)=-2.81keV$

Clearly, the ground state energy of a muonic hydrogen atom is $-2.81keV$.