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Dr. G. S. Kalyanasundaram Memorial School

Chozhan Maligai, Patteeswaram
Physics
Grade :12 Chapter-12-Atoms
[Conceptual and Basic Numerical Questions]

Chapter 12-Atoms: Alpha-particle scattering experiment, Rutherford's model of

atom, Bohr model, Energy levels, Hydrogen spectrum.

1. Alpha-particle scattering Experiment:

 An atom has a tiny positively charged core (called nucleus)which contains most
of the mass(99.9%) of the atom. The electrons move in orbits around the
nucleus because if they were at rest, they would fall into the nucleus due to
electrostatic attraction between the positive nucleus and negative charge of
electrons.
 Alpha particle 1 and 1’ which pass through the atom at a large distance from
the nucleus experience small repulsive force due to nucleus and undergo very
small deflection.
 Alpha particles 2 and 2 ‘which pass closer to the nucleus experience large
repulsive forces and hence scatters through large angles.
 Very rarely, alpha particle such as 3 travels head-on towards the nucleus.
 This alpha particle slows down due to the repulsive force of the nucleus, finally
steps and is then repelled back.

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2. Distance of closest approach (Size of nucleus):

 It can be used to estimate the size of the nucleus

 As it approaches the positive nucleus, it experiences repulsive force due to
nucleus and its kinetic energy starts getting converted into electrostatic
potential energy.

m=mass of the alpha particle V=initial velocity of the alpha particle

q1=charge on alpha particle(+2e) q2 =charge on the nucleus(+Ze)

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3. Impact Parameter:

It is defined as the perpendicular distance of the velocity vector of the alpha

particle from the centre of the nucleus, when the alpha particle is far away from the
nucleus of atom. It is denoted by b.

If value of b is large , cot(θ/2) will be large –Scattering angle will be small.

If value of b is small, cot(θ/2) will be small-Scattering angle will be large.

If b=0, cot(θ/2)=0, θ=1800-alpha particle is towards the centre of the nucleus, it

retraces its path.

4. Total Energy of an Electron:

e = charge on electron Z=total number of protons in the nucleus

m =mass of the electron r =distance of electron from the nucleus

v= linear velocity of the electron Charge on the nucleus=Ze

Force of attraction between electron and the nucleus is

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Centripetal force required to keep the electron moving in circular path

Since the atom is stable Fe = FC

Limitations of Rutherford’s Model of Atom:

i)According to Maxwell’s theory of electromagnetism ,a charge that is accelerating

radiates energy as electromagnetic waves. The electron moving around the nucleus
is under constant acceleration and therefore ,it should continuously lose
energy.Due to this continuous loss of energy ,the electrons in Rutherford’s model
were bound to spiral towards the nucleus and fall into it when all of their rotational
energy were radiated. Hence, Rutherford’s atomic model cannot be stable while in
actual practice an atom is stable. This shows that Rutherford’s model is not correct.

Ii)During inward spiralling ,the electron’s angular frequency continuously increases.

As a result, electrons will radiate electromagnetic waves of all frequencies. But this
contrary to observation. Experiments show that an atom emits line spectra and
each line corresponds to a particular frequency /wavelength.

Failed to explain the stability of the atom and the emission of line spectra.

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5. Bohr’s Model of atom:

 The electrons revolve around the nucleus of the atom in circular orbits.
 The centripetal force required by electrons for revolution is provided by the
electrostatic force of attraction between the electrons and nucleus.
 An electron can revolve only in those circular orbits in which its angular
momentum is an integral multiple of h/2π.

 While revolving in stable or stationary orbits, the electrons do not radiate

energy in spite of their acceleration towards the centre of the orbit.

 The revolving electron emits energy in the form of electromagnetic waves as

it jumps from outer stationary orbit of higher energy to the inner stationary
orbit of lower energy.

 Conversely a right amount of energy is required to lift

an electron from an inner stationary orbit to an outer
stationary orbit.

 Emission of radiation as the electron jumps from the

outer stationary orbit to the inner stationary orbit.

 If the electron is to be lifted from the lower stationary orbit of energy E1 to

a higher stationary orbit E2 , the external energy required
Equal to E2 –E1 has to be supplied to it.

6. Bohr’s Theory of Hydrogen Atom:

e =charge on electron
m = mass of electron
r n =radius of the nth orbit
Z = number of positive charges
Vn = velocity of the electron in the nth orbit

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 Radius of Bohr’s Stationary orbits:

 Velocity of an electron:

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 Frequency of Electron:

 Total Energy of an Electron:

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 Wavelength of Emitter Radiation:

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7. Spectral Series of hydrogen atom:

 Long before Bohr, many scientists had studied experimentally the spectrum
of hydrogen atom.
 Bohr gave a mathematical explanation for this spectrum.
 The whole hydrogen spectrum can be divided into distinct groups of lines,
each group of lines is called spectral series

 Lyman Series: It is obtained when electrons jump to first orbit from outer
orbit. It lies in the ultraviolet region

 Balmer Series: It is obtained when electrons jump to second orbit from outer
orbit. It lies in the Visible Spectrum.

 Paschen Series: It is obtained when electrons jump to third orbit from outer
orbit. It lies in the Infra-red Region.

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 Brackett Series: It is obtained when electrons jump to fourth orbit from outer
orbit. It lies in the Infra-red Region.

 P-fund series: It is obtained when electrons jump to fifth orbit from outer
orbit. It lies in the Infra-red Region.

8. Energy Level Diagram: It is a diagram in which the total energies of electron in

different stationary orbits of an atom are represented by parallel horizontal lines,
drawn according to some suitable energy scale.

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9. De Broglie’s Explanation of Bohr’s Postulate of Quantisation of Angular

Momentum:

 According to de-Broglie ,a particle of mass m moving with speed v is

associated with a wave of wavelength λ is given by

 Orbiting electron around the nucleus of an atom is associated with a

stationary wave.

 Since electron move in circles, the electron wave must be circular standing
wave that closes on itself.

 If the wavelength does not close upon itself, destructive interference takes
place as wave travels around the circular orbit and it quickly dies out.

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10.Excitation Energy: The minimum energy required to excite an atom in the

ground state to one of the higher stationary states.

11. Excitation Potential: The minimum accelerating potential which provides an

electron energy sufficient to jump from the ground state (n=1) to one of the outer
orbits.

12. Ionisation Energy: The minimum energy required to ionise an atom.

13. Ionisation Potential: The minimum accelerating potential which would provide
electron energy sufficient to just remove the electron from the atom.

Conceptual and Basic Numerical Questions

1. What is the distance of closest approach?

ANS : The minimum distance up to which an energetic α-particle travelling directly

towards a nucleus can reach.

2. An electron in a hydrogen atom is revolving round a positively charged nucleus.

Which two physical quantities explain the orbit of an electron?

ANS: Two Physical quantities are i) angular momentum and ii) total energy of
electron.

3. What will happen if an electron instead of revolving becomes stationary in H-atom?

ANS : Then the electrostatic field of the nucleus will attract the electron into the
nucleus itself.

4. A proton strikes another proton at rest. Assume impact-parameter to be zero; i.e.,

head on collision. How close will the incident proton go to other proton?

ANS: Change in K.E of proton= Change in P.E of the two proton system

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5. Define Impact parameter.

ANS: The perpendicular distance of the velocity vector of an alpha particle from the
central line of the nucleus, when the particle is far away from the nucleus of the
atom.

6. In a head-on collision between an alpha particle and a gold nucleus, the distance of
nearest approach is 4 X10 -14 m. Calculate the energy of the alpha particle. Atomic
number of gold =79.

ANS:

7. For scattering by an ‘inverse square law’ field (such as that produced by a charged
nucleus in Rutherford’s model),the relation between impact parameter b and the
scattering angle ϴ is given by :

a) What is the scattering angle for b=0?

b) For a given impact parameter b, does the angle of deflection increase or decrease
with increase in energy?
c) What is the impact parameter at which the scattering angle is 900 for Z=79 and
initial energy equal to 10 MeV?
d) Why is it that the mass of the nucleus does not enter the formula above but its
charge does?
e) For a given energy of the projectile, does the scattering angle increase or decrease
with decrease in impact parameter?
ANS:

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a) For b=0

b) For given b,

c)

d)The scattering of alpha particles takes place due to charge on the nucleus .If Z=0
,θ=00 .Mass of nucleus does not appear in the expression for b because the recoil of
the nucleus is being ignored i.e., the nucleus is assumed to be at rest during its
interaction with the alpha particle.

e) For a given energy (1/2 mv 2 ) of the projectile, the decrease in the value of
impact parameter means a decrease in the value of cot θ/2 and hence an increase
in the value of scattering angle θ

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8. Why do we use a very thin gold foil in Rutherford’s alpha particle scattering
experiment?

ANS : i) A very thin gold foil is used to ensure that alpha particle is deflected in a
single collision.

ii) Gold can be beaten into thin sheets.

iii) Gold nucleus is heavy and can produce large deflection of alpha particles.

9. What is the main conclusion of Rutherford’s alpha particle scattering experiment?

ANS:

i) Most of the alpha particles pass straight through the gold foil or suffer only
small deflections
ii) A small number of alpha particles suffered fairly large deflections.
iii) A very small number of alpha particles (about 1 in 8000)practically retraced
their paths or suffered deflection of nearly 1800.

10. Write the limitations of Rutherford’s Model of atom:

ANS: i) According to Maxwell’s theory of electromagnetism, a charge that is

accelerating radiates energy as electromagnetic waves. The electron moving
around the nucleus is under constant acceleration and therefore it should
continuously lose energy. Due to this continuous loss of energy, the electrons in this
model were bound to spiral towards the nucleus and fall into it when all of their
rotational energy were radiated. Rutherford’s atomic model cannot be stable while
in actual practice, an atom is stable. This shows that Rutherford’s model is not
correct.

ii)Rutherford’s model fails to explain the line spectra.

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11. Write four postulates of Bohr’s atom model:

i)An electron in an atom could revolve in certain stable orbits without the emission
of radiant energy.
ii)An electron revolves around the nucleus only in those orbits for which angular
momentum is an integral multiple of h/2π,where h is the Planck’s constant
L= m v r = n h/2π h=6.6X10-34 J s

iii)An electron might make a transition from one of its specified non- radiating
orbits to another of lower energy. Due to transition, a photon is emitted having
energy equal to the energy difference between the initial and final states. The
frequency of the emitted photon is given by h ν = Ei - Ef

12. The radius of innermost electron of a hydrogen atom is 5.3 X10 -11 m. What is the
radius of orbit in the second excited state and first excited state?

ANS:

i) n =3

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ii)n=2

13.Using Bohr’s model ,calculate the speed of the electron in a hydrogen atom in the
n=1,2 and 3 levels.

ANS:

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14.Using Bohr’s Model,calculate the orbital period of the electron in a hydrogen atom
in the n=1,2,3 levels.

ANS:

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15.Calculate the shortest wavelength of Lyman,Balmer,Paschen,Brackett and Pfund

series.

ANS:

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16.Calculate the longest wavelength of Lyman,Balmer,Paschen,Brackett and Pfund

series.

ANS:

17.Calculate the wavelength of second line of Lyman Series.

ANS:

18.Write any three limitations of Bohr’s theory:

ANS: i) It is applicable to simplest atoms like hydrogen with Z=1.It fails for other
elements.
ii)It could not explain the wave properties of electrons.
iii)It could not explain the difference in the intensities of emitted radiations.
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19.Name the series of hydrogen spectrum lying in the infrared region.

ANS: Paschen Series,Brackett Series and Pfund Series.

20. Name the series of hydrogen spectrum lying in the visible region and ultraviolet
region.

ANS: Visible Region-Balmer Series . Ultraviolet Region- Lyman Series

En

-0.85eV n=4
-1.51eV
n=3

-3.4eV n=2

-13.6eV n=1

21. The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and
potential energies of the electron in this state?

ANS: E=-13.6 e V

i) K.E = - E = 13.6 e V ii) P.E = -2 K.E = -27.2 e V

22.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 the second to the third Bohr orbit . Find the Value of Z.

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ANS:

23. Define Excitation energy:

ANS: The minimum energy required to excite an atom in the ground state to one of
the higher stationary states.

24. Define Excitation Potential:

ANS: The minimum accelerating potential which provides an electron energy

sufficient to jump from the ground state (n=1) to one of the outer orbits .

25.Define Ionisation energy.

ANS: The minimum energy required to ionise an atom .

26. Define ionisation Potential.

ANS: The minimum accelerating potential which would provide on electron energy
sufficient to just remove the electron from the atom.

27.Energy of 13.6 eV is required to ionise a hydrogen atom. What is the energy

required to remove the electron from n=2 state?

ANS:

28. The energy E of a hydrogen atom with principal quantum number n is given by

What is the energy of a photon ejected when the electron jumps

.
from n=3 state to n=1 state?
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ANS:

29.The energy of an atom are shown in figure. Which one of these transitions will
result in the emission of a photon of wave length 275 nm?

ANS:

30. The energy level diagram of an element is given in the figure .Identify ,by doing
necessary calculations which transition corresponds to the emission of a spectral line
of wavelength 102.7nm.

ANS:

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31. Name the spectral series which lies in the visible region.

ANS: Balmer series.

32. Calculate the ratio of the frequencies of the radiation emitted due to the
transition of the electron In a hydrogen atom from Its (i) second permitted energy
level to the first level and (ii) highest permitted energy level to the second permitted
level.

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33. Answer the following questions, which help you understand the difference
between Thomson’s model and Rutherford's model better.

(a) Is the average angle of deflection of α-particles by a thin gold foil predicted by
Thomson's model much less, about the same, or much greater than that predicted by
Rutherford's model?

(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles

greater than 90°) predicted by Thomson's model much less, about the same, or much
greater than that predicted by Rutherford's model?

(c) Keeping other factors fixed, it is found experimentally that for small thickness t,
the number of α-particles scattered at moderate angles is proportional to t. What
clue does this linear dependence on t provide?

(d) In which model is it completely wrong to ignore multiple scattering for the
calculation of average angle of scattering of α-particles by a thin foil?

ANS :(a) about the same

The average angle of deflection of α-particles by a thin gold foil predicted by
Thomson's model is about the same size as predicted by Rutherford's model. This is
because the average angle was taken in both models.

(b) much less

The probability of scattering of α-particles at angles greater than predicted by

Thomson's model is much less than that predicted by Rutherford's model.

(c) Scattering is mainly due to single collisions. The chances of a single collision
increases linearly with the number of target atoms. Since the number of target
atoms increase with an increase in thickness, the collision probability depends
linearly on the thickness of the target.

(d) Thomson's model

It is wrong to ignore multiple scattering in Thomson's model for the calculation of

average angle of scattering of α-particles by a thin foil. This is because a single
collision causes very little deflection in this model. Hence, the observed average
scattering angle can be explained only by considering multiple scattering

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34. The gravitational attraction between electron and proton in a hydrogen atom is
weaker than the coulomb attraction by a factor of about 10 -40 .An alternative way
of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen
atom if the electron and proton were bound by gravitational attraction. You will
find the answer interesting.
ANS:

35. The kinetic energy of alpha particle incident on a gold foil is doubled. How does
the distance of closest approach change?

ANS: The kinetic energy of alpha particle incident on a gold foil is doubled,
distance of closest approach is halved[r0 α 1/K]

-------All the Best --------

Practice NCERT Book back Exercises and Examples of all the chapters

Dr.G.S.K Memorial School Chapter -12-Atoms Department of Physics