Team Quanta gladly presents all possible short questions of Modern Physics & Electronics’ Chapter#01: Origin of Quantum Mechanics.
Q.4.1 Show that Plank’s constant has dinemsions of angular momentum. Does it mean that angular momentum is necessarily a quantized quantity?
Answer: Plank’s constant is given by relation
$$E=hv$$
$$\Rightarrow h=\frac{E}{v}$$
Dimensions of ‘h’ are $ML^2T^{-2}=ML^2T^{-1}$. Angular momentum is given by relation,
$$L=mvr$$
Dimensions of angular momentum are,
$$MLT^{-1}L=ML^2T^{-1}$$
So Plank’s constant and angular momentum have same dimensions. Angular momentum is quantized according to relation
$$L=\sqrt{l\left(l+1\right)}$$ where $\hbar=\frac{h}{2\pi}$
Q.4.2 Betelgeuse is a red giant star in the constellation Orion; Rigel is a bluish white star in the same constellation. Is the surface temperature of Betelgeuse equal to the surface temperature of Rigel?
Answer: Red light has longer wavelength than blue light, so from Wien’s displacement law, red star has lower temperature than blue star. Hence Betelgeuse has lower surface temperature than that of Rigel.
Q.4.3 Is there a lowest temperature below which black body radiation is no longer given by an object?
Answer: An object at absolute zero gives no black body radiations. But according to laws of thermodynamics, absolute zero is not attainable so all bodies with finite temperature give off black body radiation.
Q.4.4 is energy quantized in classical mechanics?
Answer: No, energy in classical mechanics is not quantized. In classical mechanics energy can have any value. If energy is quantized, then it has only discrete values.
Q.4.5 why a glass tube used to examine photoelectric emission is evacuated?
Answer: In photoelectric emission the tube used evacuated to avoid collisions of photoelectrons with air atoms or molecules.
Q.4.6 Light of given wavelength ejects electrons from surface of one metal but not from surface of another metal. Why?
Answer: All metals have different work functions. So a light of given wavelength ejects electrons from surface of one metal but not from other because second metal may have greater work function.
Q.4.7 why a tube used to examine photoelectric effect is fitted with a quartz window rather than glass?
Answer: In photoelectric effect quartz window is used because it admits only ultraviolet light while glass window, if used, will admit all the incident light spectrum.
Q.4.8 If X-rays are scattered from protons instead of electrons, is the change in their wavelength for a given angle of scattering increased?
Answer: The change in wavelength for a given angle is proportional to $\frac{h}{m_0C}$. As proton is heavier than electron, so change in wavelength will be decreased.
Q.4.9 why are photoelectric measurements so sensitive to the nature of photoelectric surface?
Answer: Photoelectric measurements are sensitive to nature of photoelectric surface because photoelectric emission is related to work function of surface.
Q.4.10 why is it that even for incident monochromatic light the photoelectrons are emitted with a spread of velocities?
Answer: Because photoelectrons are ejected only from a single orbit but from different orbits, so for incident monochromatic light, the photoelectrons are emitted with a spread of velocities.
Q.4.11 In an experiment of photoelectric effect, a beam of light with frequency greater than threshold frequency constant, does the number of electrons ejected per second from metal surface increase?
Answer: As intensity of light remain constant and frequency increase, so number of electrons ejected per second decreases.
Q.4.12 Explain the statement that “one’s eye could not detect faint star light if light were not particle like”.
Answer: One’s eye could not detect faint star light if light were not particle like because photons are scattered even by electrons with longer wavelength. This would not be the case with faint star light consisting of wave behavior.
Q.4.13 Photon A has twice the energy of photon B. What is ratio of momentum of photon A to photon B?
Answer. As photon A has twice the energy of photon B, so
$$E_{A}=2E_{B}$$
$$\Rightarrow hv_{A}=2hv_{B}$$
$$\Rightarrow \frac{hv_{A}}{c}= \frac{2hv_{B}}{c}$$
$$\Rightarrow P_{A}=2P_{B}$$
$$\Rightarrow P_{A} ∶ P_{B}=2∶1$$
Q.4.14 what is Compton wavelength? Calculate the Compton wavelength of an electron and proton.
OR
Why do not observe Compton Effect with visible light?
Answer. From Compton relation, we have
$$(1-cos \phi )$$
For $\phi =90^{ \circ }$
$\bigtriangleup \lambda = \frac{h}{m_{0}c}$ is called Compton wavelength.
So;
$$\lambda_{c} = \frac{h}{m_{0}c}$$
For electron
$$λ_{C}=\frac{6.63 \times 10^{-34}js}{9.11 \times 10^{-31} kg \times 3 \times 10^{8}m⁄s}=2.43 \times 10^{-12} m$$
For proton
$$λ_{C}=\frac{6.63 \times 10^{-34}js}{1.67 \times 10^{-27} kg \times 3 \times 10^{8}m⁄s}=1.32 \times 10^{-15} m$$
Which is smaller compared to that of electron. This explains why Compton Effect is not detected with visible light which involves particles of large dimensions for scattering.
Q.4.15 In both photoelectric effect and Compton Effect there is an incident photon and an ejected electron. What is difference between these two effects?
OR
Distinguish between photoelectric effect and Compton Effect.
Answer: Photoelectric effect:
- When ultraviolet light falls on metal surface, electrons are given out which are called photoelectric and process is called photoelectric effect.
- In photoelectric effect, the incident photon transfers all of its energy to electron.
- Photoelectric effect confirms particle like nature of photon with quantized energy.
- Photoelectric effect takes place with ultraviolet light.
Compton Effect:
- Compton Effect is a process in which an incident photon of frequency ʋ is scattered by a stationary electron and scattered photon has its frequency .
- In Compton Effect a photon transfers a part of its energy to electron and is scattered with remaining energy.
- Compton Effect confirms particle like nature of photon with quantized energy as well as momentum.
- Compton Effect cannot be observed with ultraviolet light.
Q.4.16 Does a black body at 2000K emit X-rays? Does it emit radio waves?
Answer: The Plank’s radiation law
$$R\left(\lambda\right)=\frac{2\pi h c^2}{\lambda^5}\ \frac{1}{\frac{hc}{e^{\lambda K_BT}-1}}$$
Show that an ideal black body emits radiations at all wavelengths; the spectral radiancy is equal to zero only for λ = 0 & when $\lambda \rightarrow\infty.$ So a black body at 2000K does indeed emit both X-rays and radio waves. However Rayleigh-Jeans law;
$$R\left(\lambda\right)=\frac{2\pi c K_BT}{\lambda^4}$$
Show that the spectral radiancy is very low for wavelengths much less than and much longer than a few hence such a black body emits very little in way of X-rays and radio wave.
Q.4.17 How does a photon differ from a material particle?
Answer: Difference between a photon and a material particle:-
Photon:
- A photon has no rest mass because it cannot be at rest w.r.t. any observer.
- A photon moves with velocity of light.
- Momentum of photon with frequency v is P=hʋC
Material particle:
- A material particle has always some mass and may be at rest.
- A material particle cannot move with velocity of light.
- Momentum of particle with mass ‘m’ moving with velocity is P=mv
Q.4.18 what are characteristics of cavity radiations of cavity radiator?
Answer: Characteristics of cavity radiations: The spectrum of cavity radiations from a hot body is continuous with following characteristics:
- The higher the temperature, the greater the amount of radiation emitted.
- The higher the temperature, the shorter is wavelength emitted radiating most intensely.
- Cavity radiation helps us to understand the nature of thermal radiation.
Q.4.19 If momentum of a particle with finite mass is doubled, its kinetic energy increases by a factor of four. If momentum of a photon is double, by what factor does its energy increase?
Answer: The energy and momentum of a photon are proportional to frequency, so doubling momentum of photon doubles its energy also.
Q.4.20 Give at least three characteristics of photon.
Answer: Characteristics of photons are:
- A photon can travel with speed of light.
- A photon is a wave like particle.
- Momentum of photon is hv and energy is $\frac{hv}{c}.$
Q.4.21 In photoelectric effect why does the existence of threshold frequency speak in favour of photon theory and against the wave theory?
Answer: According to photon theory, a beam of light of frequency v consists of a stream of photons, each photon having energy where ‘h’ is Plank constant. When light in form of photon of energy falls on a metal, the photons transfer their energy to electron. If frequency of incident photon is smaller than a critical value called threshold frequency, the photoelectric effect does not take place. It is established by experiments that for a given metal surface, there exits such a threshold frequency, below which light is unable to produce photoelectric effect, even though the light beam may be very intense. This fact is in accordance with the photon theory and hence we can say that the existence of threshold frequency speak in favor of photon theory.
Q.4.22 Is Compton Effect more supportive of photon theory of light than is the photoelectric effect? Explain your answer.
Answer: Compton Effect is more supportive of photon theory than the photoelectric effect because Compton Effect clearly shows the particle like nature of photon. Not only can a precise energy be assigned to a photon as in photoelectric effect but also a precise momentum .
Q.4.23 how does Compton Effect support the Einstein photon hypothesis?
Answer: From Compton relation
$$\frac{1}{v’}-\frac{1}{v}=\frac{h}{m_0c^2}\left(1-\cos{\emptyset}\right)$$
Where ʋ is frequency of incident photon and is frequency of scattered photon and is rest mass of electron. When Compton Effect is demonstrated experimentally, it is found that frequency of scattered photon is same as calculated by relation given above this relation has been established by photon theory. Hence Compton Effect supports Einstein theory photon theory.
Q.4.24 Can photoelectric effect take place with a free electron? Explain.
Answer: We know that photoelectric effect is produced when a light of threshold frequency falls on a metal surface. A free electron cannot produce photoelectric effect as its frequency may be less than the threshold frequency and at the same time electrons of metal may repel the free electrons.
Q.4.25 what is meant by threshold frequency? How can we determine the maximum kinetic energy of photon electrons?
Answer: The minimum value of frequency of incident light at which electrons are emitted from a surface is called threshold frequency.
The maximum kinetic energy of photoelectrons can be determined by reversing the polarity of plates E and C and by providing a suitable retarding potential difference to photoelectrons between plates. As potential difference is gradually increased, the current shown by ammeter goes on decreasing till for a certain value of retarding potential difference the current becomes zero, this potential difference is known as stopping potential difference or cut off potential. The maximum kinetic energy is that value with which the electrons can overcome this retarding potential difference and is given by
$$K_{max}=\frac{1}{2}mV^2=V_0e$$
Q.4.26 (a)-When ultraviolet light falls on a certain dyes, visible light is emitted. Why does this not happen when infrared light falls on these dyes?
(b)-Will bright light eject more electrons from a metal surface than dimmer light of the same color?
Answer: (a)-As frequency of ultraviolet is greater than frequency of infrared light, so visible light is emitted with ultraviolet and not with infrared.
(b)-Yes, because bright light is more intense than dimmer light of same color.
Q.4.27 In the photoelectric effect, explain why the stopping potential depends on the frequency of light but not on the intensity.
Answer: According to Plank’s quantum theory, energy of photon is given by
$$E=hv$$
This shows that the photon energy depends on the frequency of the light. The intensity gives only the number of photons reaching a unit area in a unit time.
Q.4.28 which has more energy, a photon of ultraviolet radiation or a photon of yellow light? Explain.
Answer: Ultraviolet light has shorter wavelength and higher photon energy than any wavelength of visible light.
Q.4.29 All objects radiate energy. Why, then are we not able to see all objects in a dark room?
Answer: Our eyes are cannot detect all wavelengths of electromagnetic waves. We are only able to see objects that emit or reflect electromagnetic radiation in the visible of the spectrum and not in infrared region etc.
Q.4.30 There is one property of photon which is established by the Compton Effect but cannot be established in photoelectric effect. What is it?
Answer: Photoelectric effect confirms particle like nature of photon with quantized energy. Compton Effect confirms particle like nature of photon with quantized energy as well as momentum.
Q.4.31(a)-Why the spontaneous emission of photoelectrons cannot be explained on basis of classical mechanics?
(b)-Does the every photon on a photo cathode cause emission of electrons?
Answer: (a)-According to classical mechanics, emission of photoelectrons is not spontaneous but actually photo electric emission is spontaneous emission process so it cannot be explained on basis of classical mechanics.
(b)-No, every photon incident on photo cathode may not cause emission of electrons.
Q.4.32 The classical model of blackbody radiation given by the Rayleigh-Jeans law has two major drawbacks. Identify the drawbacks and explain how Plank’s law deal with them.
Answer: According to Rayleigh-Jeans law, the intensity of short wavelength radiation emitted by black body approaches infinity as the wavelength decreases. The second failure is the prediction of much more power output from a black body than is shown experimentally.
Planck answered both failure by stating that energy of photon is not continuous as in classical physics but has discrete values and brought the theory into agreement with the experimental data by adding an exponential term to the denominator that depends on .
Q.4.33 A beam of red light and a beam of blue light have exactly the same energy. Which beam contains greater number of photons?
Answer: For photons,
$$E=nhv\ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow E=\ \frac{nhc}{\lambda}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ nhcE=\lambda\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow n\propto\lambda$$
This relation shows that light with greater wavelength has large number of photons. Therefore red light contains greater number of photons than blue light.
Q.4.34 (a)-Which photon of light red, green or blur carries more energy and momentum?
(b)-Does the brightness of beam of light primarily depend on the frequency of photons or number of photons?
Answer: (a)-Energy of photons is given by relation,
$$E=hv\ \ \ \ \ \ \ \ \ \Rightarrow E=\frac{hc}{\lambda}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow E\propto\frac{1}{\lambda}$$
As energy of photon is inversely proportional to wavelength, so photon of light having smaller wavelength has greater energy. Therefore blue light has maximum energy among red light, blue light and green light.
Moreover momentum of photon is,
$$p=\frac{h}{\lambda}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow p\propto\frac{1}{\lambda}$$
Hence same is the case with momentum.
(b)-As brightness increase with intensity of light, so brightness of beam of light primarily depends on number of photons.
Q.4.35 If the detector in the Compton experiment is moved so that scattered x-rays are detected at an angle large than ? Does the wavelength of the scattered x-rays increase or decrease as the angle u increases?
Answer: According to Compton shift in wavelength,
$$\lambda^\prime-\lambda\ \ \ =\frac{h}{m_0C}\left(1-cos\theta\right)$$
If the angle increases, decreases and hence increase. Therefore, the scattered wavelength increases.
Review Questions
R.4.1 Every metal has a definite work function. Why not photoelectrons come out, all with same energy if incident radiation is monochromatic? Why is there energy distribution of photoelectrons?
Answer: Every metal has definite work function. Work function is minimum energy that must be supplied to an electron in highest level of conduction band to come out from metal. All electrons in metals are not of same energy, so other electrons in different energy level will come out with different energies.
R.4.2 Two metals A and B have work functions 2eV and 4Ev respectively. Which metal have lower threshold wavelength for photoelectric effect?
Answer: As work function of metal B is greater than that of metal A, so metal B has greater threshold frequency than metal B. Therefore metal B has lower threshold wavelength than metal A.
R.4.3 When light from a lamp fall on a wooden table, no photoelectrons are emitted from table.
Answer: The work function of wood is greater than that of energy of incident photon so electrons cannot be ejected from the wood.
R.4.4 The distance between photocell and source of light is d. what is effect on stopping potential if distance is doubled?
Answer: The stopping potential depends on frequency of incident light. As frequency is not changing but intensity is changing, therefore stopping potential will not change.
R.4.5 If frequency of incident light on cathode of photocell is doubled, how will the following change?
- Kinetic energy of ejected electrons
- Photoelectric current
- Stopping potential
Answer: If frequency of incident light is doubled, then kinetic energy of ejected electrons will be increased twice.
The current will not change.
The stopping potential will also be greater than twice because it depends on kinetic energy.
R.4.6 Discuss the change in color for a platinum wire heated at different temperatures and what does it infer?
Answer: When platinum wire is heated, it appears dull red at about C, changes to cherry red at , becomes orange red at , yellow at and finally white at . This shows that as the temperature rises, the radiation becomes richer in shorter wavelength.
R.4.7 How the temperature affects the wavelength of radiation emitted by a black body?
Answer: With increase in temperature of a black body, the wavelength of emitted radiation decreases.
R.4.8 (a)-On what factors the radiation emitted by a black body depends?
(b)-What happens to total radiation from a black body if its absolute temperature is doubled?
Answer: (a)-The radiations emitted by a black body depends on temperature of body.
(b)-According to Stefan-Boltzmann law,
$$E=\sigma T^4$$
According to above relation when temperature is doubled, amount of radiant energy will increase 16 times.
R.4.9 If the photoelectric effect is observed for one metal, can you conclude that the effect will also be observed for another metal under the same conditions?
Answer: No, if the second metal have larger work function than the first, then the incident photons may not have enough energy to eject photoelectrons.
R.4.10 Suppose a photograph was made of a person’s face using only a few photons. Would the result be simply a very faint image of the face?
Answer: A photon can interact with the photographic film at only one point, so few photons would only give a few dots of exposure, apparently randomly scattered.
