Atom (JEE LEVEL)

Since the initial and final kinetic energy of the system of two particles remains equal hence there is no loss of kinetic energy and hence excitation of hydrogen atom is not possible in this collision

Perfectly inelastic collision

We know that in this case maximum loss of kinetic energy takes place. Lets calculate the maximum loss of kinetic energy when a neutron collide head on with a hydrogen atom.

Applying conservation of linear momentum

mv+0=mv’+mv’        

(remember \displaystyle {{m}_{{neutron}}}=\text{ }{{m}_{{hydrogen}}} = m)

\Rightarrow \text{   }mv=2mv’    

or   \Rightarrow \text{   }v’=\frac{v}{2}

Let \displaystyle \Delta E be the loss in kinetic energy then  \displaystyle \Delta E=\text{ }K{{E}_{{initial}}}\text{ }-K{{E}_{{final}}}. Remember that \displaystyle \Delta E is the energy available for excitation of atom.

where       \text{K}{{\text{E}}_{{initial}}}\text{=}\frac{1}{2}m{{v}^{2}}+0

and K{{E}_{{final}}}=\frac{1}{2}m{{{v}^{\prime}}^{2}}+\frac{1}{2}m{{{v}^{\prime}}^{2}}

therefore K{{E}_{{final}}}=\frac{1}{2}m{{\left[ {\frac{v}{2}} \right]}^{2}}+\frac{1}{2}m{{\left[ {\frac{v}{2}} \right]}^{2}}

so K{{E}_{final}}=\frac{1}{4}m{{v}^{2}}   (because \text{ }v’=\frac{v}{2})

Since \Delta E=\text{K}{{\text{E}}_{{initial}}}-K{{E}_{{final}}}

hence \Delta \text{E=}\frac{1}{2}m{{v}^{2}}-\frac{1}{4}m{{v}^{2}}

\Rightarrow \text{    }\Delta \text{E=}\frac{1}{4}m{{v}^{2}}=\frac{1}{2}K{{E}_{{initial}}} 

So we have seen that half of the kinetic energy of neutron is lost during the collision. Now this energy is absorbed by the hydrogen atom to get excited.

Minimum K.E. of neutron required to excite the hydrogen atom

To just excite a hydrogen atom from ground state i.e.n=1 to n=2 level the energy required is 10.2eV.

Hence during the collision  if \displaystyle \Delta E =10.2 eV then the hydrogen atom will get excited from n=1 to n=2 level.

Hence \displaystyle \Delta E =10.2 eV

or    \frac{1}{2}K{{E}_{{initial}}}=10.2

therefore  K{{E}_{{initial}}}=20.4eV

So to just excite a hydrogen atom from n=1 to n=2, the colliding neutron must have at least 20.4eV of energy.

To excite the hydrogen from n=1 to n=3 level how much energy must the neutron have?

Answer is that fro transition from n=1 to n=3 the energy required is  13.6-1.5= 12.1eV.

So loss in energy i.e. \displaystyle \Delta E must be at least equal to 12.1eV

Hence \displaystyle \Delta E =12.1eV

or   \frac{1}{2}K{{E}_{{initial}}}=12.1

therefore K{{E}_{{initial}}}=24.2eV.

So to excite a hydrogen atom from n=1 to n=3, the colliding neutron must have at least 24.2eV of energy.

Important Point:

                          Just now we have seen that to excite the atom from n=1 to n=2 level the loss of energy \displaystyle \Delta E must be 10.2eV. And for n=1 to n=3, \displaystyle \Delta E must be 12.1eV. Now what happens if \displaystyle \Delta E is lying between the two values like if \displaystyle \Delta E is 11.5eV?

                          Answer is in that case the atom will absorb 10.2eV and get excited from n=1 to n=2, whereas the remaining energy will remain with the particles as their kinetic energies. Hence the collision is not perfectly inelastic but it can be said as partially inelastic

Illustration  

The longest wavelength in the Lyman series for hydrogen is 1215 \displaystyle {{A}^{o}}. Calculate the Rydberg constant.

Solution      

\frac{1}{\lambda }=R\ \left( {\frac{1}{{n_{l}^{2}}}-\frac{1}{{n_{u}^{2}}}} \right)

For the Lyman series \displaystyle {{n}_{l}} = 1; the longest wavelength will correspond to the value \displaystyle {{n}_{u}} = 2.

\frac{1}{{1215{\AA}}}=R\ \left( {\frac{1}{{{{1}^{2}}}}-\frac{1}{{{{2}^{2}}}}} \right)  

or R = \displaystyle 1.097\times {{10}^{{-3}}}{{{\AA}}^{{-1}}}

Illustration

How many different photons can be emitted by hydrogen atoms that undergo transitions to the ground state from the n = 5 state?

Solution      

Consider the problem for arbitrary n. If \displaystyle {{n}_{u}}>{{n}_{l}} is any pair of unequal integers in the range 1 to n, it is clear that there is at least one route from state n down to the ground state that includes the transition \displaystyle {{n}_{u}}\to {{n}_{l}}. Thus, the number of photons is equal to the number of such pairs, which is \left( {_{2}^{n}} \right)=\frac{{n\ (n-1)}}{2}

For n = 5, there are 5(4)/2 = 10 photons.

Illustration

A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the third Bohr orbit.

 (a)  Find the value of Z

 (b) Find the energy required to excite the electron from n = 3 to n = 4

(c)  Find the wavelength of radiation required to remove electron from first Bohr’s Orbit to infinity.

(d) Find the kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.

 Solution

 (a)  Given \displaystyle \Delta {{E}_{{23}}}= 47.2 eV

Since   \displaystyle \Delta E=\text{ }13.6 \times {{Z}^{2}} \left( {\frac{1}{{n_{1}^{2}}}-\frac{1}{{n_{2}^{2}}}} \right) eV

or 47.2 = \displaystyle 13.6 \times  {{Z}^{2}} \left( {\frac{1}{{{{2}^{2}}}}-\frac{1}{{{{3}^{2}}}}} \right)

hence     Z = 5

(b) To find \displaystyle \Delta {{E}_{{34}}}, put    \displaystyle {{n}_{1}} = 3 and \displaystyle {{n}_{2}} = 4

\displaystyle \Delta E=\text{ }13.6 \times {{Z}^{2}} \left( {\frac{1}{{n_{1}^{2}}}-\frac{1}{{n_{2}^{2}}}} \right) eV

or {\Delta}E= \displaystyle 13.6 \times {{{5}^{2}}} \left( {\frac{1}{{{{3}^{2}}}}-\frac{1}{{{{4}^{2}}}}} \right)

hence  {\Delta}E = 16.53 eV

(c)  Ionization energy is the energy required to excite the electron from n = 1 to n = \displaystyle \infty

Thus,   {\Delta}E= \displaystyle 13.6 \times  {{{5}^{2}}}   \left( {\frac{1}{{{{1}^{2}}}}-\frac{1}{{{{\infty }^{2}}}}} \right)

so  {\Delta}E= 340 eV

The respective wavelength is

\displaystyle \lambda = \frac{{hc}}{{\Delta E}}= \frac{{12400}}{{\Delta E}}=\frac{{12400}}{{340}}

hence  \displaystyle \lambda = 36.47 {{A}^{o}}

(d)    we know for an electron , according to Bohr’s Theory, kinetic energy K, Potential energy U and Total energy E are related as

K= – E and  U = 2E

Hence K = –E = +340 eV

and U = 2E = -680 eV

we know that angular momentum is given by L = \frac{nh}{{2\pi }}

hence L= \frac{{1 \times 6.63\times {{{10}}^{{-34}}}}}{{2\pi }}

so  L = \displaystyle 1.056\times {{10}^{{-34}}}J-s

Illustration

Find the quantum number n corresponding to excited state of He⁺ ion if on transition to the ground state, the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm. The ionization energy of H atom is 13.6 eV.

Solution                     

The energy transitions for the given wavelengths are

\displaystyle \Delta {{E}_{1}} = \frac{{12400}}{{{{\lambda }_{1}}}}=\frac{{12400}}{{1085}}=11.43eV

\displaystyle \Delta {{E}_{2}} = \frac{{12400}}{{{{\lambda }_{2}}}}=\frac{{12400}}{{304}}\,

\displaystyle \Delta {{E}_{2}}= 40.79 eV

Total energy emitted \displaystyle \Delta E=\Delta {{E}_{1}}+\Delta {{E}_{2}} = 52.22 eV

we know that energy emitted is given by

\displaystyle \Delta E=\text{ }13.6{{Z}^{2}} \left( {\frac{1}{{n_{1}^{2}}}-\frac{1}{{n_{2}^{2}}}} \right)  eV               

or    \displaystyle 52.22 = 13.6 \times {{5}^{2}}   \left( {\frac{1}{{{{1}^{2}}}}-\frac{1}{{{{n}^{2}}}}} \right)

Thus,   n = 5

Illustration

An isolated hydrogen atom emits a photon of 10.2 eV.

(a)  Determine the momentum of photon emitted

(b) Calculate the recoil momentum of the atom

(c)  Find the kinetic energy of the recoil atom.

[Given mass of proton, \displaystyle {{m}_{P}}=\text{ }1.67\times {{10}^{{-27}}}kg]

Solution

(a)  Momentum of the photon is 

{{p}_{1}} = \frac{E}{c} = \frac{{10.2\times 1.6\times {{{10}}^{{-19}}}}}{{3\times {{{10}}^{8}}}}

thus  {{p}_{1}} = 5.44 \times {{10}^{-27}}  kg m/s

(b) Applying the momentum conservation

\displaystyle {{p}_{2}}={{p}_{1}}=5.44\times {{10}^{{-27}}}kg m/s

(c)  K = \frac{1}{2}m{{v}^{2}}  (v = recoil speed of atom, m = mass of hydrogen atom)

or K = \frac{1}{2}m{{\left( {\frac{p}{m}} \right)}^{2}}=\frac{{{{p}^{2}}}}{{2m}}

Substituting the value of the momentum of atom, we get

K = \frac{{{{{\left( {5.44\times {{{10}}^{{-27}}}} \right)}}^{2}}}}{{2\times 1.67\times {{{10}}^{{-27}}}}}

thus K = \displaystyle 8.86\times {{10}^{{-27}}}J

Illustration

A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV. Find the energy and wavelength of photon emitted.

Solution

Since the binding energy is always negative, therefore,

\displaystyle {{E}_{i}} = -0.85 eV

Let \displaystyle {{n}_{i}} be the initial binding state of the electron, then

\displaystyle {{E}_{n}} = -13.6\frac{{{{Z}^{2}}}}{{n_{i}^{2}}}

or        -0.85 = -13.6 \frac{{{{Z}^{2}}}}{{n_{i}^{2}}}

or        \displaystyle {{n}_{i}} = 4

we know that Binding energy = \displaystyle {{E}_{n}}=\text{ }-13.6\text{ }{{Z}^{2}}/{{n}^{2}}

hence -0.85 eV = -13.6 \displaystyle {{\left( 1 \right)}^{2}}/{{n}_{2}}^{2}~~  

thus  \displaystyle {{n}_{2}}  = 4

Let the electron now goes to an energy level n whose excitation energy is 10.2 eV. Since the excitation energy \displaystyle \Delta E is defined with respect to ground state, therefore

\displaystyle \Delta E=\text{ }13.6{{Z}^{2}} \left( {\frac{1}{{n_{1}^{2}}}-\frac{1}{{n_{2}^{2}}}} \right) eV

or   10.2 = \displaystyle 13.6 \times {{\left( 1 \right)}^{2}} \left( {\frac{1}{{{{1}^{2}}}}-\frac{1}{{n_{f}^{2}}}} \right)

thus     \displaystyle {{n}_{f}} = 2

So the electron makes a transition from energy level \displaystyle {{n}_{i}} = 4 to \displaystyle {{n}_{f}} = 2.

Thus, the energy released is \displaystyle \Delta E={{E}_{4}}-{{E}_{2}}

or {\Delta}E = 13.6 \times \left[ {\frac{1}{{{{2}^{2}}}}-\frac{1}{{{{4}^{2}}}}} \right] = 2.55 eV 

Since  \displaystyle \lambda= \frac{{hc}}{{\Delta E}}

hence \displaystyle \lambda =  \frac{{12400}}{{2.25eV}}=5511 {{A}^{o}} 

Illustration

A particle of charge equal to that of an electron, –e and mass 208 times the mass of electron (called a \displaystyle \mu -meson) moves in a circular orbit around a nucleus of charge +3e (take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system:

(i)   Calculate the radius of nth Bohr orbit

(ii)  Find the value of n, for which the radius of orbit is approximately the same as that of first Bohr orbit for the hydrogen atom;

(iii) Find the wavelength of radiation emitted when the m-meson jumps from the third orbit to first orbit

(Rydberg’s constant = \displaystyle 1.097\times {{10}^{7}}/m)

Solution

 (i)   Radius of the nth Bohr orbit for hydrogen atom is

\displaystyle {{r}_{n}}= 0.53\times \frac{{{{n}^{2}}}}{Z}

Since   r \propto \frac{1}{m}

Radius of nth orbit for \displaystyle \mu -meson is

\displaystyle {{r}_{n}}  = \frac{{0.53{{n}^{2}}}}{{\left( {208} \right)Z}}

or    \displaystyle \displaystyle {{r}_{n}}= (8.5\times {{10}^{{-4}}}){{n}^{2}}

(ii)    \displaystyle (8.5\times {{10}^{{-4}}}){{n}^{2}}= 0.53

  or \displaystyle {{n}^{2}} = 623

   hence  n » 25

(iii) In case of hydrogen like atom,

\displaystyle \Delta E={{E}_{3}}-{{E}_{1}}=\text{ }13.6{{Z}^{2}} \left( {1-\frac{1}{{{{3}^{2}}}}} \right)

so  \displaystyle \Delta E= 12.08 eV

since    {\Delta}E \displaystyle \propto m

so for \displaystyle \mu -meson, {\Delta}E= (12.8)(208) = 22.6 keV

Since    \displaystyle \lambda = \frac{{12400}}{{\Delta E}}=\frac{{12400}}{{22.6\times {{{10}}^{3}}}}\,

so  \displaystyle \lambda= 0.548  {{A}^{o}}