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Sunday, 8 February 2015

F.Sc ICS Notes: Physics XII: Chapter 19 Dawn of Modern Physics Exercise Short Questions:

FSc ICS Notes: Physics XII: Chapter 19 Dawn of Modern Physics Exercise Short Questions:

Question 19.1 What are measurements on which two observers in relative motion always agree upon?
Answer 19.1 The measurements of
  • force
  • acceleration
  • velocity of light
by two observers in different inertial frames of reference will be identical.


Question 19.2 Does the dilation means that time really passes more slowly in moving system or that it only seems to pass more slowly?
Answer 19.2 The time dilation is purely due to relative motion. It really happen when observers are in relative motion. The result is applied to the timing process physical, chemical and biological. Even aging process of the human body is slowed down by motion at very high speed. It should be noted that we can not detects such changes in daily life because we are not moving at speed comparable to the speed of the light.


Question 19.3 If you are moving in a spaceship at a very high speed relative to the Earth, would you notice a difference (a) in your pulse rate (b) in the pulse rate of people on Earth?
Answer 19.3 The pulse rate of the person in spaceship is the proper pulse rate. There will be no difference in the pulse rate as the inertial frame of reference is same. Pulse rate of the people on the earth will be low because they are moving with respect to spaceship.
 
 
Question 19.4 If the speed of light were infinite, what would the equations of special theory of relativity reduce to?
Answer 19.4 When the speed of light approached to infinity.
            m = m, l = l, t = t , c = ∞.
            m = mo / (√ 1 – v2 /∞) = mo / (√ 1 – 0) = mo.
            l   = lo (√ 1 – v2 /∞) = lo (√ 1 – 0) = lo.
            t   = to / (√ 1 – v2 /∞)  = to / (√ 1 – 0)  = to.
            E = m (∞)Square ⇒ E is infinity.



Question 19.5 Since mass is a form of energy, can we conclude that a compressed spring has more mass than the same spring when it is not compressed?
Answer 19.5  From classical point of view the mass remain same. According to theory of relativity, change of mass is due to relative motion and not due to position.


Question 19.6 As a solid is heated and begins to glow, why does it first appear red?
Answer 19.6 According to wien's displacement law at low temperature the solid will emit radiation having longer wavelength which lies in the infra-red region of electromagnetic spectrum. That is why the appearance of the body will be red when it begins to glow. On further increasing the temperature, the solid will change the color towards decreasing wavelength of visible spectrum.


Question 19.7 What happens to total radiation from a black body if its absolute temperature doubled?
Answer 19.7  According to Stefen Boltzmann law, the amount of energy radiated per sec per unit area pf the black body is directly proportional to the fourth power of its absolute temperature i.e.
                                                       E = σ T4
            It is obvious that if temperature T is doubled, E becomes 16 times of the original radiant energy.



Question 19.8 A beam of red light and a beam of blue light have exactly the same energy. Which beam contains the greater number of photons?
Answer 19.8  The energy of the photon of light by Planck quantum theory is given by:
                                                    E = hf
             Where h is the Planck constant and f is the frequency of light. Let n1 be the number of photons of red light having frequency f1 and n2 be the number of photons of blue light having frequency f2. The respective energies of red and blue beams of light is given be:
                                          E1 = n1 h f1
                                          E2 = n2 h f2
                                          E1 = E2
                                   n1 h f1 = n2 h f2
                                     f1 / f2 = n1 / n2
                                           f1 < f2
                                          n2 < n1
            i.e. number of photons of red light are more than the number of photons of blue light.


Question 19.9 Which photon, red, green, or blue carries the most (a) energy and (b) momentum?
Answer 19.9 The expression for energy and momentum of photons are given by
                                      E = hf
                                      p = hf / c
                                 as  f  = c / λ
                      therefore  E = hc / λ
                              and  p = h / λ
             Blue light has smaller wavelength than red and green colors. So energy and momentum of photons of Blue light is greater than green and red.



Question 19.10 Which has the lower energy quanta? Radio waves or X-rays.
Answer 19.10 By Plank quantum theory the energy of the quanta of radiation is given by,
                                    E = hf
              Since the frequency of x-rays is greater than the frequency of radio wave therefore, the quanta of x-ray will carry more energy than that of radio wave.


Question 19.11 Does the brightness of a beam of light primarily depends on the frequency of photons or on the number of photons?
Answer 19.11 Brightness or intensity of light beam of specific color means the number of photons of light passing per sec per unit area. So, brightness depends upon numbers of photons and not on frequency of photons.


Question 19.12 When ultraviolet light falls on certain dyes, visible light is emitted. Why does this not happen when infrared light falls on these dyes?
Answer 19.12 The ultraviolet light contains photons of high energy. When they fall on the atoms of the dye, the atoms get excited. On De-excitation they emit photons of frequency which lie in visible region of electro-magnetic spectrum. However infra-red light photon has less energy, so the atoms may be excited, on De-excitation they emit photons whose frequency lie in the invisible region.


Question 19.13 Will bright light eject more electrons from a metal surface than dimmer light of the same color?
Answer 19.13 The number of photo-electrons emitted from the metal surface is proportional to the intensity of light. So, bright light eject more electrons from metal surface than the dimmer light of same color.


Question 19.14 Will higher frequency light eject greater number of electrons than low frequency light?
Answer 19.14 The number of photo electrons emitted from the metal surface is directly proportional to the intensity of incident light and not on its frequency. The light of higher frequency light will eject energetic electrons, while low frequency light will eject less energetic electrons. So light of higher and lower frequency of same intensity will eject same number of electrons.


Question 19.15 When light shines on a surface, is momentum transferred to the metal surface?
Answer 19.15 When light falls on metal surface, the photons are absorbed by the surface. Therefore both energy and momentum of the photons are transferred to the atoms of the surface. The exchange of energy and momentum to the surface is so small that it hardly produce any disturbance in photo emissive surface.


Question 19.16 Why can red light be used in a photographic dark room when developing films. blue or white light cannot?
Answer 19.16 Photons of red light in visible spectrum has the longest wavelength and the least energy. Therefore photographic films and materials concerned are least affected in the presence of red light than blue and white light.


Question 19.17 Photon A has twice the energy of photon B. What is the ratio of the momentum of A to that of B?
Answer 19.17  As we know that:
                                      EA = hfA = hc / λA
                          and       EB = hc / λB
                        since       EA = 2EB
                                hc / λA = 2(hc / λB)
                                     λB = 2λA
                             momentum of photon B is given by:
                                     pB = h / λB
                        putting   λB = 2λA
                                     pB = h / 2λA
                                    as h / λA = pA (momentum of photon A)
                                     pB = 0.5pA
                                  or pA = 2pB
                                pA / pB = 2 / 1
                                pA : pB = 2 : 1                        
      


Question 19.18 Why don’t we observe a Compton effect with visible light?
Answer 19.18 Photons of visible light (violet) are most energetic and have energy:
                              E = hc / λ = {6.6 * 10(-34) x 3 * 10(8)} / 400 * 10(-9) J
                              E = {6.6 * 10(-34) x 3 * 10(8)} / 400 * 10(-9) x 1 / 1.6 * 10(-19) eV
                              E = .310 KeV
But for Compton effect energy more than .1 MeV is needed, so Compton effect cannot be studied with visible light.


Question 19.19 Can pair production take place in vacuum? Explain.
Answer 19.19 Pair production takes place near the nucleus which takes recoil to conserve momentum. Thus law of conservation of momentum will be violated if pair production takes place in vacuum. Hence pair production cannot take place in vacuum as it does not contain matter.
 

Question 19.20 Is it possible to create a single electron from energy? Explain.
Answer 19.20 No, it is not possible, because law of conservation of charge will be violated when single electron will be created.


Question 19.21 If electrons behaved only like particles, what pattern would you expect on the screen after the electrons passes through the double slit?
Answer 19.21 When electrons have particles like property diffraction will not take place and electrons would pass straight through the slits. Therefore the diffraction pattern on the screen will not be observed. We will only observe the images of two slits on the screen.


Question 19.22 If an electron and a proton have the same de Broglie wavelength, which particle has greater speed?
Answer 19.22 As we know that momentum:
                                    p = mv = h / λ
                                or v = h / mλ
                as same λ & h , and λ being constant
                                      v ∝ 1 / m
From above equation it follows that electron being lighter particle will have greater speed than proton to have same value of de Broglie wavelength.


Question 19.23 We do not notice the de Broglie wavelength for a pitched cricket ball. Explain why?
Answer 19.23 According to de Broglie, the wavelength of the particle of mass m when moving woth velcity v is given by:
                       λ = h / mv
It follows that wavelength λ is inversely proportional to the mass of the particle. The cricket ball is very much massive, so it has very short wavelength which is not measurable.


Question 19.24 If the following particles all have the same energy, which has the shortest wavelength? Electron, alpha particle, neutron, proton.
Answer 19.24  From the relation:
                                  λ = h / mv
for same energy (or K.E.) v & h are constant, so
                                  λ ∝ 1 / m
It follows from the above relation that for greater mass there is  shorter wavelength. As α-particle has greatest mass, so the α-particle has the shortest wavelength and the electron of smallest mass has the largest wavelength associated with it.


Question 19.25 When does light behave as a wave? When does it behave as a particle?
Answer 19.25 In case of interference and diffraction phenomena light behaves as waves. In phenomena like photoelectric and Compton effects, the light exhibits its particle nature.


Question 19.26 What advantages an electron microscope has over an optical microscope?
Answer 19.26 The electron microscope has the following advantages over optical microscope.
  1. The wavelength associated with electron is much shorter than the visible light, so electron microscope has high resolving power as compare to optical microscope.
  2. The energy of electrons is sufficiently large that they can penetrate through the thicker parts of matter as compared to visible light. Therefore the internal structure of an object can be obtained by Electron microscope.


Question 19.27 If measurements show a precise position for an electron, can those measurements show precise momentum also? Explain.
Answer 19.27  No. According to Uncertainty principle, position and momentum of a particle cannot both be measured simultaneously with perfect accuracy. For a precise position of an electron, the momentum becomes uncertain.
                                  Δx • Δp ≈ h
                 If the position of an electron is precisely measured then there is not uncertainty in position i.e.,
                                   Δx = 0,
                                  0 • Δp ≈ h
                                   or Δp ≈ h / 0 ≈ ∞
                 So, if measurement show a precise position for electron in an experiment then precise measurement of momentum of an electron is impossible in that experiment.

Written by: Asad Hussain.

3 comments:

  1. question 19.24!! how can you say that "For same energy, velocity is constant for all particles??"

    ReplyDelete