Bihar Board - Class 12 Physics - Chapter 12: Atoms Long Answer Question
Long Answer Type Questions
1. Using de Broglie’s hypothesis, explained with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
Answer:
According to de Broglie’s hypothesis.
λ = hmv ……………………….. (i)
According to de Broglie’s condition of stationary orbits, the stationary orbits are those which contain complete de Broglie wavelengths.
2πr = nλ ………………………….. (2)
Substituting value of λ from eq. (2) in eq. (1), we get
2πr = nhmv
⇒ mvr = n h2 ………………………… (3)
This is Bohr’s postulate of quantisation of energy levels.
2. In an experiment of α-particle scattering by a thin foil of gold, draw a plot showing the number of particles scattered versus the scattering angle θ. Why is it that a very small fraction of the particles are scattered at θ > 90°?
Answer:
A small fraction of the alpha particles scattered at angle θ > 90° is due to the reason. If impact parameter ‘b’ reduces to zero, coulomb force increases, hence alpha particles are scattered at angle θ>9O°, and only one alpha particle is scattered at angle 180°.
3. (i) State Bohr postulate of hydrogen atom that gives the relationship for the frequency of emitted photon in a transition,
(ii) An electron jumps from fourth to first orbit in an atom. How many maximum numbers of spectral lines can be emitted by the atom? To which series these lines correspond?
Answer:
(i) Bohr’s Third Postulate: It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, 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
hv = Ei– Ef
Where Ei and Ef are the energies of the initial and final states and Ei > Ef .
(ii) Electron jumps from fourth to first orbit in an atom
∴ Maximum number of spectral lines can be
4C2 = 4！2！2！ = 432 = 6
The line responds to Lyman series (e– jumps to 1st orbit), Balmer series (e– jumps to 2nd orbit), Paschen series (e– jumps to 3rd orbit).
4. The ionization energy of a hydrogen atom is 13.6 eV. An electron which jumps from ground state to n = 4 state. Calculate the wavelength of a photon.
Answer:
Energy in ground state of hydrogen atom E1 = -136 eV
Energy in n = 4 state E4 = - 13.642 = – 0.85 eV
∴ Energy of photon = E4 – E1= – 0.85 + 13.6 = 12.75 eV
∵ hν = E and c = νλ
∴ hc=E
or =hcE= 6.6210-34310812.751.610-19
⇒ λ = 973 x 10-10 m = 973 Å.
5. State main postulates of Bohr’s atomic model.
Answer:
The main postulates of Bohr’s atomic model are as follows :
1. Entire mass and entire positive charge of the atom is concentrated at the nucleus.
2. The electrostatic attraction force between nucleus and electron provides the required centripetal force to electrons to revolve around the nucleus.
3. The electron does not revolve around the nucleus in all possible orbits, but they revolve in some specific orbits. These orbits are called stable orbits.
4. The stable orbits are those orbits in which the angular momentum of electron is integral multiple of h2, where h is Plank constant (h = 6.63 x 10-34 J-sec.).
If m is mass of electron r is radius of orbit and v is its speed then angular momentum
mvr = nh2 , where n = 1, 2, 3,
n is called the principal quantum number.
6. What should be the value of angle of scattering for impact parameter b = 0?
Answer:
We know that
Impact parameter b = Ze2cot(/2)4012mv2
If b = 0, then cot(θ/2) = 0 = cot(90°)
⇒ 2 = 90°
or θ = 180°
7. Describe Rutherford’s experiment related to α – particles scattering. Also write conditions drawn from it.
Answer:
The experimental arrangement of Rutherford’s α – particles scattering experiment is shown in fig. A radioactive substance called ‘Po’ is kept in a lead box from the hole ‘O’ the α – particles are emitted out with high speed. After passing through D1 and D2 these α – particles are incident on gold leaf in the form of a beam
The thickness of the gold foil is about 10 cm. The scattered α – particles fall on screen S in which ZnS is coated α – particles produce a flash of light on this screen which can be observed by a microscope M.
Conclusion:
Most of the α – particles pass through the foil straight undeflected. Thus it can be concluded that most of the atoms are hollow.
α – particles are repelled, so there should be positive charge on the nucleus.
The entire positive charge should be on the center of the atom (or nucleus).
Importance:
By Rutherford's experiment it has come to be known that the positive charge of the atom is concentrated at the central core of the atom, which is called the nucleus.
8. If the value of energy for an impact parameter is increased then, will the scattering angle increase or decrease?
Answer:
Impact parameter b = Ze2cot(/2)4012mv2
= Ze2cot(/2)40E
If the energy is increased then in the same ratio, the value of cot(θ/2)should be increased i.e., the angle of scattering will decrease.
9. Draw a schematic arrangement of the Geiger-Marsden experiment. How did the scattering of α-particles by a thin foil of gold provide an important way to determine an upper limit on the size of the nucleus? Explain briefly.
Answer:
(i) Beams of α-particles get deviated at various angles with different probabilities.
(ii) α-particles with least impact parameter suffer larger scattering – rebounding on head-on collision.
(iii) For larger impact parameters, the particle remains almost undeviated.
Explanation:
The fact that the number of incident particles rebounding back is only a small fraction, means that the number of α-particles head-on collision is small. This implies that the entire positive charge of the atom is concentrated in a small volume. This confirms that the nucleus of the atom has an upper size limit.
10.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?
Answer: (a) The average angle of deflection (θ) of a-particles by a thin gold foil as predicted by Thomson's model is about the same that predicted by Rutherford's atom model.
(b) The probability of backward scattering of a-particles as predicted by Thomson's model is much less than that predicted by Rutherford's model.
(c) Linear dependence of the number of a-particles scattered at moderate angles on thickness t of gold foil suggests that scattering is predominantly due to a single collision. Chances of a single collision increase linearly with the number of target atoms and hence, linearly with the thickness of gold foil.
(d) In Thomson model, positive charge is uniformly distributed in the entire atom. So, a single collision causes very little deflection and hence, average scattering can be explained only on the basis of multiple scattering. Therefore, it is completely wrong to ignore multiple scattering in Thomson model. In the Rutherford model, the whole positive charge is concentrated in the nucleus and most of the deflection comes from a single scattering. So, multiple scattering may be ignored.
11. The fission properties of 94Pu239 are remarkably identical to those of 92U235. The average energy emitted per fission is 180 MeV. How much energy in MeV is discharged if all the atoms in 1kg of pure 94Pu239 experience fission?
Answer: Given that,
Average energy emitted per fission of 94Pu239, Eav=180 MeV
Amount of pure 94Pu239, m = 1kg = 1000g
NA= Avogadro number = 6.023×1023
Mass number of 94Pu239 = 239g
1 mole of 94Pu239 contains NA atoms.
Therefore, 1kg of94Pu239 contains {( NA / Mass number)×m} atoms
⇒ {(6.023×1023) / 239} ×1000 = 2.52×1024 atoms
Thus, the total energy released during the fission of 1 kg of 94Pu239: E = Eav×2.52×1024
⇒E = 180×2.52×1024
= 4.536 × 1026 MeV
Therefore, 4.536 × 1026 MeV is released if all the atoms in 1 kg of pure94Pu239 undergo fission.
12. State Bohr’s postulates for an atom ?
Answer: Bohr’s atom model : Using Maxwell’s planck quantum theory, Both explained the spectrum of Hydrogen atoms and presented a new model of atom. It has following postulates:
(i) Atoms are spherical whose central part is positively charged called the nucleus. The charge on the nucleus is +ze, where z is atomic number & e electric charge.
(ii) Electrons move around the nucleus in circular orbits & they do not emit energy. They are called non radiating paths or stable orbits.
(iii) The electrostatic force between nucleus & electron provides the necessary centripetal force i.e…
mv2r=140zeer2
.
(iv) Electrons revolve in those orbits only in which their angular momentum is integral multiple of h/2π
mvr = nh2
.
(v) When electron goes from higher orbit to lower orbit, there is emission of energy but when it goes from lower to higher orbit, energy is absorbed in terms of photon (hv) i.e
E2-E1=h .
Only a photon of fixed frequency is emitted from an atom due to which the line spectrum is obtained.
13. The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-11 m. What is the radius of orbit in the second excited state?
Answer: Radius (r) = n2 × 5.3 × 10-11m
Radius of second excited state (n =3) is:
r = (3)2 × 5.3 × 10-11 m
= 9 × 5.3 × 10-11 m
= 4.77 × 10-10 m
14. Answer the following question, which will help you better understand the difference between Thomson's and Rutherford's models.
- 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?
- 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?
- 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?
- In which model is it completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of α-particles by a thin foil?
Answer:
(a) About the same;
The average angle of deflection of α-particles by a delicate gold foil indicated by Thomson's model is about the same size as Rutherford's model. And this is because the average angle was accepted in both these models.
(b) Much less;
The possibility of diffusion of α-particles at angles greater than 90 degrees predicted by Thomson's model is much less than Rutherford's model.
(c) We know that diffusion is large because of single crashes. The probabilities of a single wreck increase linearly with the number of target atoms. Since the number of target atoms increases with an increase in thickness, the collision probability would depend linearly on the thickness of the target.
(d) Thomson's model;
It is wrong to ignore multiple scattering in Thomson's model to calculate the average angle of dispersion of α-particles by a thin foil. A single collision could cause very little deflection in this model. Hence, you can observe the average scattering angle and explain it only by considering multiple scattering.
15. Calculate the height of the potential barrier for a head-on collision of two deuterons. (Hint: The size of the possible wall is given by the Coulomb repulsion between the two deuterons when they touch each other. Assume that they can be taken as hard spheres of radius 2.0fm.)
Answer: If two deuterons collide head-on then, the distance between their centers is:
d = Radius of 1st deuteron + Radius of 2nd deuteron
Radius of a deuteron nucleus = 2 fm
= 2 × 10-15 m
⇒ d = 2 × 10-15 + 2 × 10-15
= 4 × 10-15 m
Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10-19 C
Potential energy of the two-deuteron system:
V = e2 / 4π0d
Where,
0 is the Permittivity of free space
1/4π0 = 9 × 109Mm2c-2
⇒ V = [ 9 × 10 × (1.6 × 10-19)2 / 4 × 10-15 ] J
⇒ V = [ 9 × 10 × (1.6 × 10-19)2 / 4 ×10-15 × (1.6×10-19) ] eV
⇒ V = 360 keV
Therefore, the height of the potential barrier of the two-deuteron system is 360 keV.
16. Explain a nuclear reactor in detail.
Answer: Nuclear reactor is a device in which nuclear fission is maintained as a self-supporting yet controller chain reaction.
Construction : Components of a nuclear reactor are shown in figure.
Main components of the reactor :
(i) Fissionable material (Fuel) : the fissionable material used in the reactor is called the fuel of the reactor. Uranium (U235), Thorium (th235) and Plutonium isotopes (Pu259, Pu240 and Pu241) are the most commonly used fuels in the reactor.
(ii) Moderator : Moderator is used to slow down the fast moving neutrons. Most commonly used moderators are graphite, water and heavy water. When heavy water is used as a moderator, then ordinary or non-enriched uranium can be used as a fuel because heavy water has more neutrons to produce fission. In case of ordinary water as moderator (heaving few neutrons), enriched uranium is used as a fuel.
(iii) Contro, Material : Control material is used to control the chain reaction and to maintain a stable rate of reaction. This material controls the number of neutrons available for fission. For example, cadmium rods are inserted into the core of the reactor because they can absorb neutrons. The neutrons available for fission are controlled by moving the cadmium rods in or out of the core of the reactor.
(iv) Colant : It is a cooling material which removes the heat generated due to fission in the reactor. Commonly used coolant are water, CO2 nitrogen etc.
(v) Protective shield : A protective shield in the form of a concrete thick wall surrounds the core of the reactor to save the persons working around the reactor from the hazardous radiations.
Working : A few 92U235 nuclei undergo fission liberating fast neutrons. These fast neutrons are slowed down to an energy of about 0.025 eV by the surrounding moderator (graphite). The cadmium rods are used to control the chain reaction. The fission produces heat in the nuclear reactor core, the coolant transfers this heat from the core to the heat exchanger, where steam produced at a very high pressure runs a turbine and the electricity is obtained at the generator. The dead steam from the turbine condenses into water and is returned to the heat exchange.
17. What fraction of tritium will remain after 25 years? Given the half-life of tritium as 12.5 years.
Answer: It is given that,
t = 25 years
T = 12.5 years
N/N0 = (1/2)tT
= (1/2)2512.5
⇒ N/N0 = (1/2)2=1/4
⇒ N/N0 = 0.25,
Which is the required fraction.
18. A difference of 2.3eV separates two energy levels in an atom. What is the radiation frequency when the atom transitions from the upper level to the lower level?
Answer: We are given the separation of two energy levels in an atom,
E = 2.3 eV
= 2.3 × 1.6 × 10-19
= 3.68 × 10-19 J
If v is the frequency of radiation emitted when the atom transits from the upper level to the lower level, the energy is given by,
E = hv
Where, h= Planck's constant = 6.62× 10-4 Js
⇒ v = E / h
Substituting the given values,
ν = 3.68 × 10-19 / 6.62× 10-32
= 6.62 × 10-32
= 5.55 × 1014 Hz
Hence, the frequency of the radiation is found to be 5.55 × 1014 Hz.
19. (a) Why only gold leaf is taken for the scattering of α – particles?
(b) The spectrum of a hydrogen atom has many lines though a hydrogen atom contains only one electron why?
Answer:
(a) For the scattering of α – particles, the leaf should be very thin, so that α – particles get scattered after collision. Moreover the nucleus should be heavy, so that the α – particles are scattered by large angles. Gold has both the properties moreover this foil of gold can easily be made.
(b) Large number of spectral lines are present in the hydrogen atom spectrum because in the light source of hydrogen a large number of atoms are there. There are different transitions in different atoms. That is why the hydrogen spectrum contains a number of lines.
20. The total energy of an electron in the first excited state of the hydrogen atom is about −3.4eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
Answer:
(a) We are given,
Total energy of the electron, E = −3.4 eV
The kinetic energy of the electron is equal to the negative of the total energy.
⇒ K.E = −E
∴ K.E = −(−3.4) = + 3.4eV
Hence, the kinetic energy of the electron in the given state is found to be +3.4eV.
(b) We know that the potential energy (U) of the electron is found to be equal to the negative of twice its kinetic energy.
⇒ U = − 2 K.E
∴ U = − 2 × 3.4
= −6.8 eV
Hence, the potential energy of the electron in the given state is found to be −6.8 eV
(c) We understand that the potential energy would rely on the connection point brought. Here, the potential energy of the connection point is considered zero. The system's potential energy value would also change by adjusting the reference point. Since we know that total energy is the sum of kinetic and potential energies, the system's total energy will also change.
21. Describe Rutherfor’s atomic model.
Answer:
Rutherford’s model of atom : Rutherford presented a model of atom in 1911 to explain the results obtained from the α-scattering experiment. This is called Rutherford’s model. Its main characteristics are as follow :
(i) Total mass and total positive charge of an atom is concentrated at its center (nucleus). The nucleus is of the order of 10-15 meters.
(ii) Electrons are distributed in a hollow sphere of radius 10-10 meter around the nucleus, the total negative charge of electrons is equal to total positive charge.
(iii) Electrons in atoms are not stationary. If they are taken to be stationary then they should fall in the nucleus due to force of attraction. But it is not so. He proposed that electrons revolve in different orbits around the nucleus. The necessary centripetal force is obtained or provided by Coulomb’s force acting between nucleus and electron.
22. The total energy of an electron in the first excited state of a hydrogen atom is −3.4eV. Calculate:
(a) K.E. of the electron in this state.
(b) P.E. of the electron in this state.
(c) Which of the answers would change if zero of PE is changed? Justify your answer?
Answer:
(a) We know that K.E = −E
∴ K.E = 3.4 eV
(b) P.E = 2 × K.E
∴ P.E = 2 × 3.4 = 6.8 eV
(c) If the zero of the P.E is changed, K.E will remain intact, but the P.E will alter, and so will the total energy.
23. (a) What does the negative energy of an electron signify?
(b) What are the stationary orbits of an electron in bohr’s atomic model?
Answer:
(a) The negative energy of an electron signifies that the electron is attracted to the nucleus. This is due to the electrostatic force of attraction between electrons and protons of the nucleus, the Potential energy is negative and also greater than the Kinetic energy of electrons. Which results in the total energy of electrons to be negative. so, It cannot escape from the atom.
(b) According to Bohr's postulate, the electrons can revolve around the nucleus in specific discrete, non-radiating orbits such that the angular momentum of an electron should be the integral multiple of $h / 2 \pi$. Such orbits are known as stationary orbits.
24. What is Rutherford’s a-scattering experiment ?
Answer:
Ratherford’s α-scattering experiment : To find out the structure of atoms Rutherford performed an experiment. In this experiment he obtained an α-particle from a radioactive substance and made an incident on atoms of different elements. He found that α-particles are deflected in different directions. He made incident α-rays on thin gold foil. the fine pencil of ray (particle) having high kinetic energy (K.E.) obtained from radioactive element polonium. To measure the deflection of α-particles a scintillation counter is used having a fluorescent screen. When α-particles strike the screen, they are visible in a microscope. Counter is adjusted in different situations and the deflection of a-particles are measured. The total arrangement is kept in vacuum, so that there is no collision of α-particles with air particles.
From this experiment Rutherford found following :
(i) Most α-particles cross gold foil without deviation. It was concluded that most of the atoms are hollow.
(ii) Some α-particles are scattered in different directions from their original path. This angular distribution of a-particles is fixed Since α-particles are positively charged so an object which is deviating them must have positive charge. It was concluded that the total positive charge of the atom is concentrated at the center.
(iii) Some α-particles (one in 20,000) are such which are returned by 90° or more. This is called back scattering. It was concluded that high repulsive force should be needed to return a-particles of high speeds.
On the basis of his experiment he gave following important facts about structure of atom:
(i) Discovery of nucleus : Rutherford said that total positive charge and mass are concentrated in a given small region called nucleus of atom. Beside it, he estimated the radius of the nucleus as 10-14 meters (order). Nucleus is 1/ 104 times the size of an atom.
(ii) Coulomb’s law is correct for atomic distance. and (iii) Calculate the size of the nucleus.
25. Calculate the kinetic energy and potential energy of an electron in the first orbit of a hydrogen atom. Given e = 1.6× 10-19C and r = 0.53× 10-10m.
Answer:
(a) Kinetic Energy
K.E.= ke2/2r
⇒K.E.=(1.6× 10-19)2×9× 109/ 2×0.53× 10-10
⇒K.E.=21.74× 10-19J
⇒K.E.=21.74× 10-19 / 1.6× 10-19 = 13.59eV
(b) Potential Energy
P.E. = −ke2 / r = −2K.E.
⇒ P.E. = −2×13.59=−27.18eV
Therefore, Kinetic energy is 13.59eV and Potential energy is −27.18eV.
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