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Life is like riding a bicycle.  To keep your balance you must keep moving.”
Albert Einstein

Wednesday, January 26, 2011

Answer to Free Response Practice Question on Atomic Physics and Quantum Effects for AP Physics B

“I know this world is ruled by infinite intelligence. Everything that surrounds us everything that exists proves that there are infinite laws behind it. There can be no denying this fact. It is mathematical in its precision.”

– Thomas A. Edison

A free response practice question involving on atomic physics and quantum effects was given to you in the post dated 23rd January 2011. As promised, I give below a model answer along with the question:

An electron and its antiparticle positron, moving with the same speed, undergo a head-on collision and get annihilated to produce two photons of equal wave length. This process is called pair-annihilation. Positron has the same mass as the electron, but its charge is positive.

(a) Determine the rest energy of a positron in electron volt.

(b) Determine the maximum wave length of the photons generated in the electron-positron pair annihilation.

(c) Which one of the following statements [(i), (ii) and (iii)] is correct regarding the direction of motion of the two photons? Put a tick (√) mark against the correct choice.

(i) The photons move along the same direction……..

(ii) The photons move along opposite directions……..

(iii) The photons move along perpendicular directions……..

(d) Explain why two photons (and not a single one) have to be produced in pair annihilation.

(e) If the particles undergoing pair annihilation were proton and its antiparticle antiproton (instead of electron and positron), what can you say about the maximum wave length (λ1) of the photons generated, in comparison with the value obtained in the case of electron-positron pair annihilation? Put a tick (√) mark against the correct answer choosing from the options (i), (ii) and (iii) given below:

(i) λ1 is increased…….

(ii) λ1 is the same…….

(iii) λ1 is decreased…….

(a) If the rest mass of the positron is m0 and the speed of light in free space is c, the rest energy E of the positron, as given by Einstein’s mass-energy relation, is

E = m0c2 = (9.11×10– 31 kg)(3×108 ms–1)2 = 8.20×10– 14 J

[The mass of the positron is the same as that of the electron]

Since one electron volt is equivalent to 1.6×10– 19 J, the energy in eV is given by

E = (8.20×10– 14 J)/(1.6×10– 19 J) = 5.12×105 eV.

(b) The wave length of the photon generated in pair annihilation will be maximum if the electron and positron have minimum energy which is equal to the rest energy. Therefore, the photon of maximum wave length has energy equal to 8.20×10– 14 J.

Since E = hc/λ where h is Planck’s constant and λ is the wave length of the photon, we have

λ = hc/E

Therefore, maximum wave length, λmax = (6.63×10– 34 Js×3×108 ms–1)/(8.20×10– 14 J)

Or, λmax = 2.43×10–12 m

(c) Statement (ii) is correct.

The photons have to move along opposite directions since the total momentum of the two photon system must be zero. This follows from the law of conservation of momentum. The total momentum of the electron-positron system was zero since they moved with the same speed in opposite directions

[Note that the magnitudes of the momenta of the photons generated are the same since they have the same wave length λ (Remember de Broglie relation p = h/λ where p is the momentum].

(d) A photon can never be at rest and hence it has momentum. The condition of zero total momentum for the system can be satisfied only if there are two photons (at least) so that the momentum of one can be nullified by that of the other.

(e) The correct choice is (iii).

The rest mass of the proton (or anti-proton) is much greater than that of the electron (or positron) and hence the rest energy is greater. The energy of the generated photon is greater and so the wave length is smaller.

You can access all posts related to atomic physics and quantum effects on this site by clicking on the label ‘atomic physics’ below this post.

Sunday, January 23, 2011

Atomic Physics and Quantum Effects for AP Physics B - Free Response Question for Practice

Today I’ll give you a free response practice question from the section, ‘atomic physics and quantum effects’ on the hope that you will try to answer it and will become aware of the use of mastering fundamental concepts in this section. Here is the question:

An electron and its antiparticle positron, moving with the same speed, undergo a head-on collision and get annihilated to produce two photons of equal wave length. This process is called pair-annihilation. Positron has the same mass as the electron, but its charge is positive.

(a) Determine the rest energy of a positron in electron volt.

(b) Determine the maximum wave length of the photons generated in the electron-positron pair annihilation.

(c) Which one of the following statements [(i), (ii) and (iii)] is correct regarding the direction of motion of the two photons? Put a tick (√) mark against the correct choice.

(i) The photons move along the same direction……..

(ii) The photons move along opposite directions……..

(iii) The photons move along perpendicular directions……..

(d) Explain why two photons (and not a single one) have to be produced in pair annihilation.

(e) If the particles undergoing pair annihilation were proton and its antiparticle antiproton (instead of electron and positron), what can you say about the maximum wave length (λ1) of the photons generated, in comparison with the value obtained in the case of electron-positron pair annihilation? Put a tick (√) mark against the correct answer choosing from the options (i), (ii) and (iii) given below:

(i) λ1 is increased…….

(ii) λ1 is the same…….

(iii) λ1 is decreased…….

I’ll post a model answer for your benefit soon. Meanwhile, answer the above question yourself so that you will be aware of your strength and weakness (if any)!

The above question carries 10 points and you may take 11 minutes for answering it.

Sunday, January 9, 2011

Multiple Choice Practice Questions on Atomic Physics and Quantum Effects for AP Physics B

There is no substitute for hard work

– Thomas A. Edison

Questions on Atomic Physics and Quantum Effects with solution were posted earlier on this site. You can access all the posts by clicking on the labelatomic physicsbelow this post or by trying a search using the search box provided on this page. Today we will discuss a few more practice questions (MCQ) in this section: (1) The adjoining figure shows the energy level diagram of a hypothetical atom. A gaseous sample containing these atoms in the ground state is irradiated with photons in the band of energies from 7 eV to 9 eV. Consequently the excited atoms in the gas will emit photons of the following energies:

(a) All energies in the band from 7 eV to 9 eV

(b) Energies 6 eV, 10 eV and 15 eV

(c) Infinite number of discreet energies from 0 eV to 10 eV

(d) Energies 4 eV, 5 eV and 9 eV

(e) Infinite number of discreet energies from 0 eV to 15 eV

The atoms in the ground state (of energy – 15 eV) can absorb 9 eV from the incident photons and can get excited to the state of energy – 6 eV since it is an allowed state for the atom as per the energy level diagram given in the question. Once in the excited state of energy – 6 eV, the atom can undergo transition to the lower states of energy – 10 eV and – 15 eV in three ways:

(i) Jump first from – 6 eV level to – 10 eV level, emitting a 4 eV photon.

(ii) Jump next from – 10 eV level to – 15 eV level, emitting a 5 eV photon.

(iii) Jump directly from – 6 eV level to – 15 eV level, emitting a 9 eV photon.

Therefore the excited gas will emit photons of energies 4 eV, 5 eV and 9 eV as given in option (d).

(2) If the gaseous sample of atoms in the above question were irradiated with photons of two discreet energies 2 eV and 12 eV, what will be the energies of the photons emitted by the gas?

(a) Energies 7 eV and 12 eV

(b) All energies in the band from 7 eV to 12 eV

(c) Energies 3 eV, 4 eV, 5 eV, 7 eV, 9 eV, and 12 eV

(d) Energies 3 eV, 4 eV and 5 eV

(e) Infinite number of discreet energies from 0 eV to 12 eV

Incident photons of energy 2 eV cannot be absorbed by the atoms since there is no energy level which is higher by 2 eV with respect to the ground level. But the incident photons of energy 12 eV can be absorbed by the atoms since there is the energy level of – 3 eV which is higher by 12 eV with respect to the ground level of energy – 15 eV.

After getting excited to the level of – 3 eV, the atom can go to the ground state in six different ways:

(i) Jump first from – 3 eV level to – 6 eV level, emitting a 3 eV photon.

(ii) Jump next from – 6 eV level to – 10 eV level, emitting a 4 eV photon.

(iii) Jump next from – 10 eV level to – 15 eV level, emitting a 5 eV photon

(iv) The atom (actually the electron in the atom) can jump from – 3 eV level to – 10 eV level, emitting a 7 eV photon.

(v) The atom, after reaching – 6 eV level [as given in (i)] can jump directly to – 15 eV level, emitting a 9 eV photon.

(vi) The atom can jump directly from – 3 eV level to – 15 eV level, emitting a 12 eV photon.

Therefore the excited gas will emit photons of energies 3 eV, 4 eV, 5 eV, 7 eV, 9 eV, and 12 eV as given in option (c).

(3) The wave property of particles is evident in

(a) scattering of photons by electrons at rest

(b) Coolidge X-ray tube

(c) electron microscope

(d) scattering of α-particles by atomic nuclei

(e) photo electric cells

The correct option is (c). In an electron microscope electron beam is used (similar to light beam in optical microscope) because of the wave property of electrons. The wave length of electron at a few hundred volts of accelerating voltage is of the order of X-ray wave length so that much greater resolution and magnification compared to optical microscope is possible.

(4) In a hydrogen atom in the ground state suppose the orbital speed of the electron is ‘v’. If the mass of the electron is ‘m’ and the magnitude of its charge is ‘e’, what is the radius of the orbit of the electron?

(a) e2/4πε0mv2

(b) e2/4πε0mv

(c) e/4πε0mv2

(d) mv2/4πε0e4

(e) mv2/4πε0e2

The centripetal force required for the circular motion of the electron is supplied by the electrostatic attractive force between the electron (of charge – e) and the nucleus (which is a proton of charge + e). Therefore we have

mv2/r = (1/4πε0) (e2/r2) where ε0 is the permittivity of free space.

This gives r = e2/4πε0mv2

Saturday, January 1, 2011

“If you think in terms of a year, plant a seed; if in terms of ten years, plant trees; if in terms of 100 years, teach the people”

- Confucius

“Happy New Year 2011”