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

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.

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.
   Example:-
   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.

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.

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.

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 . λ_red >
λ_blue, so beam of red light will contain a greater number of photons.

(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.

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.

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.

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.

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.

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.

Yes, When light shines on metal surface, both momentum and energy are transferred to electrons of the surface.

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.

1. Energy of photon A = E_A = hf_A
   Energy of photon B = E_B = ½ hf_A       (E=hf)
   Or   E_B = ½ E_A…………………………………(1)

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

Now
Momentum of photon A = p_A = E_A/c………………….(2)
Momentum of photon B = p_B = E_B/c
But from equ (1) we can write it
P_B = E_B/c  =        1/2E_{A  ………………………………………………..} (3)
c
Dividing equ (2) by equ (3) we get
P_A/P_B = E_A/c * c/1/2E_A = 1/1/2 = 2
or p √ p_B = 2
Hence, momentum of photon A is also two times greater than the momentum of photon B.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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  = hf_o
The relation shows that greater the work function, greater will be the threshold frequency ‘f_o’. Hence copper will have higher threshold 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.

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.

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.

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.

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.

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.

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.

(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.

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.

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.

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.

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

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.

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.

Photoelectric equations can be written as
kE_max = hf- φ
Where hf = incident energy
φ   = work function
K.E_max = Maximum K.E of photoelectrons

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.

A positron is an anti- particle of electron. Thus, the positron has the same mass that at of electron but it carries positive charge.

Photocell are used to operate
(i). Automatic door system
(ii). Sound strake of movies
(iii). Security system
(iv). Automatic street light
(v). Counting system

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