12th Class Physics Chapter 8 DAWAN AND MODREN PHYSICS Short Question Answers

Physics short QA

12th Class Physics Chapter 8 DAWAN AND MODREN PHYSICS Short Question Answers Below

1. What are the measurements on which two observers in relative motion will always agree upon?
The measurements on which the two observers in uniform relative motion (i.e moving with uniform motion w.r.t each other) will always agree upon are speed of light in free space, magnitudes of force and acceleration of a moving object.
2. Does the dilation mean that time really passes more slowly in moving system or that it only seems to pass the slowly?
3. If you are moving in a spaceship at a very high speed relative to the Earth, would you notice a difference (a) I your plus rate (b) in the plus rate of people on the Earth?
4. If the speed of light were infinite, what should be the equations of special theory of relativity reduce to?
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?

  1. Yes, there would be an increase in the mass of compressed spring. However, this increase in mass is slightly greater than the mass of uncompressed spring due to condensation of energy of compressed spring into mass (i. e work done in compressing the spring has changed into energy which increase the mass of the spring) However, this increase in mass is neglibly small.
    Suppose a spring is compressed and energy stored in it is 10 j. Then by using the relation
    ΔE = Δmc2
    or   Δm = ΔE/c2 = 10/ (3*108)2   = 10/ 9*1016
    or Δm   = 1.1 *10-15 kg
    This increase in mass is negligible as compared to the original mass of spring.

6. As a solid is heated and being to glow, why does it first appear red?
When a body is heated, it emits radiations. The nature of radiation depends upon the temperature. At low temperature a body emits radions of long appears red. As the temperature of the solid is further increased, the proportion of radiations of short wavelength increases. This is why the solids when heated look red first.
7. What happens to total radiations from a blackbody of its absolute temperature is doubled?
According to Stefan – Boltzman,s law,
E =  T4
When the temperature is increased to double of its value, then
E/  = σ (2T)4
Or E/ = σ * 16 T4
E/ = 16σ T4 = 16E
Thus, the total radiation energy E/ will increase 16 times.
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?
As we know that energy of single photon is given by
E = hf
Let n be the number of photons, then the energy becomes
E = nh f …………………….(i)
But c = fλ  or  f  = c/λ
Putting the value of f in equ (1) we have
E = nhc/λ
or n = Eλ/hc…………………………(2)
As a red and blue light have the same energies, so E, h and c have constant values. Thus
n  = constant * λ
or        nλ …………………………(3)
The relation (3) shows that greater wavelength will have a large number of photons.
But wavelength of red light is greater than that of blue light i.e . λred >
λblue, so beam of red light will contain a greater number of photons.
9. Which photon, red, green, or blue carries the most (a) energy and (b) momentum?

(a). As we know that energy of the photons given by

E = hf
But            f= c/λ
E = hc/λ………………………..(1)      (c =fλ)
But      h and c have constant value. Therefore
E = constant   * 1/λ
or E  1/λ

This relations shows, smaller the wavelength, larger the energy. As the wavelength of photon of blue light is shorter among red and green light photon, so
(a). Photon of blue light has maximum energy.
(b). As momentum of photon is given by

P = h/λ
But h is constant, therefore
P = cont * 1/λ
or p  1/λ
According to this relation, wavelength of blue light is shorter among red and green light photons. So
(b). Photon of blue light has maximum momentum.

10.Which has the lower energy quanta? Radio waves or X rays?
As frefuency of X – rays is greater than that of radio waves or wavelength of X – rays is smaller than that of radio waves, so E ∝ f (or 1/λ) This relation shows that energy of X – ray quantum is greater than that of radio waves. It means that the radio wave has the lower energy quantum.
11.Does the brightness of a beam of light primarily depends on the frequency of photons or on the number of photons?
When the light is visible or not, it depends upon the frequency. But the brightness of a beam of light depends upon the number of photons and not on the frequency of light or photons. Thus, the brightness increases with the increase in intensity of light, i.e. due to increase in number of photons.
12.When ultraviolet light falls on certain dyes, visible light is emitted. Why does this not happen when infrared light falls on these dyes?
The ultraviolet light has photons of high energy. When they fall on certain dyes, they can excite the atoms of dyes. When the excited atoms return to their ground state ( or de – excited), they emit frequencies which are detectable by normal human eye. However, when infrared light is made to fall on the atoms of dyes, the they may be excited. But the de – excitations takes place in such a way that photons of frequencies are emitted whose values are below the least value of frequencies are emitted spectrum. Thus, they cannot be detected by normal eye. Hence, the ultraviolet light falling on dyes causes the light visible while the infrared light cannot do so.
13. Will bright light eject more electrons from a metal surface than dimmer light of the same color?
Yes, as we have studied in the phenomenon of photo – electric effect, the number of electrons emitted from the surface of metal is directly proportional to the intensity of light. Therefore, bright light being more of the same color and having less intensity.
14. Will higher frequency light eject greater number of electrons than low frequency light?
No, the number of photoelectrons does not depends upon the frequency but depends upon the intensity of light. Therefore, high frequency light will not emit more electrons than a low frequency light. It means that both high and low frequency lights will emit the same number of electrons.
15. When light shines on a surface, is momentum transferred to the metal surface?
Yes, When light shines on metal surface, both momentum and energy are transferred to electrons of the surface.
16. Why can red light be used in a photographic dark room when developing films but a blue or white light cannot?
Since the frequency of red light is smaller than that of blue and white lights, so red light has less energy as compared to blue and white light. Therefore it (red light) cannot affect the photographic plate and make no preferred in a photographic dark room. While blue and white lights having more energy that of red light can affect the photographic plates, and cause chemical reactions.
17. Photon A has twice the energy of photon B. What is the ratio of the momentum of A to that of B?

  1. Energy of photon A = EA = hfA
    Energy of photon B = EB = ½ hfA       (E=hf)
    Or   EB = ½ EA…………………………………(1)

As we know that momentum of photon is given by
p = h/λ = hf /c = E/c                  (c = fλ)
or λ = c/f

Momentum of photon A = pA = EA/c………………….(2)
Momentum of photon B = pB = EB/c
But from equ (1) we can write it
PB = EB/c  =        1/2EA  ……………………………………………….. (3)
Dividing equ (2) by equ (3) we get
PA/PB = EA/c * c/1/2EA = 1/1/2 = 2
or p √ pB = 2
Hence, momentum of photon A is also two times greater than the momentum of photon B.

18. Why don’t we observe a Compton Effect with visible light?
Compton effect is observed with X –ray and λ- ray photons. These photons have energy which is thousand times greater visible lights photons. During Compton Effect, an X – ray photon knocks an electron out of the atom. Even after this interaction the X – ray photon has sufficient energy so that it retains its identity. On the other hand, visible light photon loses all of its energy in a single interaction with an electron in an atom. Therefore, we cannot observe a Compton Effect with visible light.
19. Can pair production take place in vacuum? Explain.
No, Pair production cannot take place in vacuum. During the process of pairs production the law of conservation of energy and law of conservation of momentum are observed. In order to conserve the momentum and energy, the presence of heavy nucleus is essential. Because pair production take place near the nucleus which take the recoil to conserve the momentum. Therefore, pair production cannot take place in vacuum because vacuum has no particle to stop the photon.
20. Is it possible to create a single electron from energy? Explain.
No, it is not possible to create a signal electron form energy. In the phenomenon of pair production, when a photon of energy hf greater than 2m0c2 strikes a nucleus, an electron – positron pair is created. A positron is an anti – particle of an electron. It is also called a positive electron. Hence, creation of electron will be against the law of conservation of charge which cannot be violated.
21. If electrons behaved only like particles, what pattern would you except on the screen after the electrons pass through the double slit?
When electrons behave only like particles, then they only would pass through either of the two slits straight and strikes the screen only just in front of the slits to produce exact images of the double slits on the screen. No interference pattern would be seen.
22. If an electron and a proton have the same de Broglie wavelength, which particle has greayter speed?
According to de- Broglie, we know that wavelength associated with a particle is given by
λ = h/ mv
or    v = h/mλ …………………………………(1)
As  de Broglie wavelength λ for both the particles Is same end ‘h’ is also constant, so equ (1) can be written as
v = constant * 1/ m
or    v    1/m
It shows that greater the mass of the particle, smaller will be its velocity. As mass of proton is greater than mass of the electron, so speed of electron would be greater than the speed of proton.
23. We do not notice the de Broglie wavelength for a pitched cricket ball. Explain Why?
According to de – Broglie relation
λ = h/ mv
Due to large mass of cricket ball and small speed. The value of wavelength ‘λ’ associated with a moving cricket ball is extremely small. So it cannot show the wave – propertmy and no diffraction can be produced. Hence, we cannot notice the de Broglie wave length for pitched cricket ball. In other words, wave length of wave associated with ball cannot be detected.
24. If the following particles all have small energy, which has the shortest wave length? Explain, alpha particle, neutron and proton.
25.When does light behave as a wave? When does it behave as a particle?
Light behave is a wave in the phenomenon of interference, diffraction and polarization. Secondly, Light behaves as a particle in photoelectric effect and Compton effect and pair production.
26. What advantages an electronic microscope has over an optical microscope?
The resoling power of an electronic microscope is thousand times greater than that of an optical microscope. Therefore, such minor details which cannot be seen by an optical microscope can be observed by an electronic microscope. A 50 kV electron microscope can resolve a distance of 0.5 nm to 1 nm optical microscope has the resolving power of 0.2 m. This is the major advantage of electron microscope over optical microscope. In an electron microscope, focusing of invisible electron beam can be done by the electron and magnetic field instead of optical lens in an optical instrument. 3. The picture of internal structure of an object can be obtained with the help of electron microscope while an optical microscope is unable to do so.
27. Measurements show a precise position for an electron, can those measurements shows precise momentum also? Explain.
No, According to Heisenberg uncertainty principle (i. e. Δp, Δx ≈h) is not possible to make precise (accurate) measurement of position and momentum of an electron simultaneously. If one quantity is precise, then the other is uncertain (with error). It is clear from the following relation Δx = h/Δp It is shows that less is uncertainty in the measurement of position of a particle, the uncertainty in the measurement of momentum is large. Hence, it is not possible to have more precise measurement of momentum when its position is measured accurately.
28.Suppose you are standing at the platform and a train passes by you. A passenger sitting in the train and you are looking at a clock on the train. Who measures the proper time interval? You or the passenger? Justify tour answer.
29. It is harder to remove a free electron from copper than sodium. Which metal has greater work function? Which metal higher threshold frequency?

The minimum energy required to break the free electron from the metal surface is called the work function ‘φ’. According to the definition of work function, copper has greater work function  than sodium.

Threshold Frequency:-

The relation between the work function and threshold frequency is given by
φ  = hfo
The relation shows that greater the work function, greater will be the threshold frequency ‘fo’. Hence copper will have higher threshold frequency.

30. Photo – electric emission occurs only for a definite frequency of light, What this particular frequency is called? Will the number of electrons emitted per second depend n the intensity of light if its frequency is greater than the particular frequency?
Photo. Electron emission takes place only at a certain definite frequency ‘f0’ of light each material. This particular frequency is called threshold frequency. If the incident light is of a single frequency greater than the threshold frequency of the metarial, the number of the electrons emitted per second is directly proportional to the intensity of light. Hence, the number of electrons emitted per second depends upon the intensity of light. It means that the number of electrons emitted per second increases with the increase of frequency above the threshold frequency.
31. If we keep on applying force on a material object, can it gain the speed of light?
32. Why must the rest mass of a photon be zero?
33. It is possible to create a single electron from energy? Explain,
No, It is not possible to create a single electron from energy. In the phenomenon of pair production when a photon of energy hf greater than 2m0c2 strikes a nucleus, an electron – positron pair is created. A positron is an antiparticle of an electron. It is called as a positive electron. As photon is electrically neutral, so in order to converse the charge an electron with a positron must be produced.
34. Why is photo – electric effect so sensitive to the nature of the metal surface?
Photoelectric effect depends upon the nature of the metal surface. Electrons of a metal require a minimum amount of energy to leaves its surface. This minimum amount of energy is known as the work function of that metal and denoted by φ. The photons of the light used to produced photoelectric effect must have energy at least equal to the work function φ of the metal. Since work function is different for different metals, therefore, the photoelectric effect depends upon the nature of the metals surface.
35. Can an instant beam of television waves focused on a metal plate cause photo – electric emission? Explain.
Television waves and photoelectric emission:- The television waves consist of photons which have frequency f which is much less than the threshold frequency f0 for common metals As the photons can only produce photoelectric emission when their frequency is at least equal to f0, therefore, an intense beam of television waves cannot cause photoelectric emission.

36. Can photo – electric effect takes place with a free electron?
photo- electric effect with a free electron:- We know that photoelectric effect is produced when a light of certain frequency I,e threshold frequency. Falls on a metal surface. A free electron cannot produce phot – electric effect as its frequency may be les the the threshold frequency and the same time electrons of the metal may repel the free electron.
37. State the postulates of special theory of relativity?
Special theory of relativity is based upon the following two postulates. (1). All the laws of physics are the same in all inertial frames of references. (2). The speed of light in free space has the same value for all observers, regardless of their state of motion.

38. What is black body?
A black body is a solid black having a hollow cavity within it. It has a small hole at its one face from where radiations can eneter or escape through this hole. A perfect hole body can emit as well as absorb all possible kinds of electromagnetic radiations.
39. Describe the important results concluded from Einstein’s special theory of relativity?
(1). When the body approaches the speed of light, the mass of the body becomes infinite. (2). When the speed of the body is comparable to the speed of light, the light of the body along the direction of motion of body decreases. (3). If the speed of the clock is comparable to the speed of light, then this running clock will run slower than an ordinary clock.
40. Give the max. Plank’s explanation of the radiations emitted by a black body?
According to Max plank, the radiations from a black body are always emitted or absorbed in the form of packets of energies. According to Max – plank the radiations from a black body are always emitted or absorbed in the form of packets of energies. Secondly, the energy associated to radiations is directly proportional to its frequency i.e. E ∝ f or E = hf Where h is plank’s constant.
41. What is photoelectric effect?
The emission of electrons from a metal surface when light of suitable frequency falls on it. This phenomenon is called photoelectric effect. The emitted electrons by this effect are called photoelectrons.
42. Would it be more convenient to demonstrate photoelectric effect with ultra – violet or visible light? Explain.
Since ultra – violet light has short wavelength than that of visible light. Therefore, it will have greater frequency as compared to visible light. Thus, a photon of ultra – violet light has greater amount of energy. Hence, it is convenient to demonstrate the phenomenon of photoelectric effect with ultra – violet light.
43. If E and P are the energy and momentum, calculate the velocity of light?
As we know the relation for energy and frequency
E = hf
Or E = hc/λ …………………………………(1)      (f = c/f)
Since  P = h/λ ……………………………..(2)
So using equs (i) and (ii) we get
E  = P.C
Or    0  = F/P
44. What do you understand by work function and stopping potential (cut off potential)?
Work function:- The minimum energy required to eject an electron from any metallic surface is called its work function. The work function is different for different metals. It is generally denoted by φ. Stopping potential (or cut off potential) . The external voltage used to sto object photoelectrons is called cut – off potential or stopping potential.
45. Write some factors upon which photoelectric effect depends?
Photo electric effect depends on the following factors. (i). The frequency of incident light. (ii). Nature of the metallic surface. (iii). Threshold frequency of the metal.
46. Write the Einstein’s photoelectric equation?
Photoelectric equations can be written as
kEmax = hf- φ
Where hf = incident energy
φ   = work function
K.Emax = Maximum K.E of photoelectrons
47. Describe some important results of the photo –electric effect?
It has been found experiments that (i). No photo- electrons are emitted when the frequency of light is below the threshold frequency. (ii). The speed of photoelectrons increases with the increase in frequency of the incident light. (iii). The number of photo- electrons emitted is directly proportional to the intensity of the incident light. (iv). The threshold frequency depends upon the nature of the metal. (V). A beam of light of frequency slightly greater than the threshold frequency, however weak and may be, causes an immediate emission of electrons.
48. What is a positron?
A positron is an anti- particle of electron. Thus, the positron has the same mass that at of electron but it carries positive charge.
49. Describe some important use of photocell?
Photocell are used to operate
(i). Automatic door system
(ii). Sound strake of movies
(iii). Security system
(iv). Automatic street light
(v). Counting system
50. What is the Campton shift in the wavelength of a photon scattered at an angle of 900?
Compton shift is given by the relation
Δλ  = h/m0c (1- cosφ)
Where m0 = 9.11 *1031 kg
h = 6.63 * 10-34js
θ = 900
c = 3*108ms-1
Δλ =           6.63*10-34                 (1-cos 900)
9.1 * 10-31 * 3*108
Δλ = 2.34 *10-12m

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