Team Quanta gladly presents all possible short questions of BS Physics book Quantum Mechanics-I’s Chapter#1: Origin of Quantum Mechanics.
Q.1.1 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
$$L=\sqrt{l\left(l+1\right)}\hbar , \ \ \ where\ \ \ \ \hbar= \frac{h}{2\pi}$$
Q.1.2 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.1.3 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.1.4 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.1.5 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.1.6 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.1.7 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.1.8 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
$$E_A=2E_B$$
$$\Rightarrow hv_A=2hv_B$$
$$\Rightarrow\frac{hv_A}{c}=\frac{2hv_B}{c}$$
$$\Rightarrow P_A=2P_B$$
$$\Rightarrow P_A \colon P_B=2\colon1$$
Q.1.9 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.1.10 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
$$\Delta \lambda =\frac{h}{m^oc}(1-cos\phi )$$
For $\emptyset={90}^0$
$\Delta \lambda =\frac{h}{m^oc}$ is called Compton wavelength.
So;
$$\lambda_C=\ \frac{h}{m_0c}$$
For electron
$$\lambda_C=\frac{6.63\times{10}^{-34}\ Js}{9.11\times{10}^{-31}kg\times3\times{10}^8\frac{m}{s}}=2.43\times{10}^{-12}m$$
For proton
$$\lambda_C=\ \frac{6.63\times{10}^{-34}Js}{1.67\times{10}^{-27}kg\times3\times{10}^8\frac{m}{s}}=1.32\times{10}^{-15}m$$
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.1.11 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=hʋ/C
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
Q.1.12 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 λ 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.1.13 what are characteristics of cavity radiations of cavity radiations?
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.1.14 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.1.15 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.1.16 how does Compton Effect support the Einstein photon hypothesis?
Answer: From Compton relation
$$1v’-1v=\frac{h}{m^oc^2}(1-cos\emptyset)$$
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.1.17 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.1.18 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.1.19 why is the wave nature of matter not more apparent in our daily observations?
Answer: De-Broglie relation is,
$$\lambda=\frac{h}{mV}$$
Above equation shows why we do not observe the wave behavior of ordinary objects. Plank’s constant ‘h’ is so small that the wavelength of ordinary objects is many orders of magnitude smaller than the size of a nucleus. No double slit could possibly be constructed on this scale to reveal the wave nature.
Q.1.20 what common expression can be used for momentum of either a photon or a particle?
Answer: The common expression which can be used for the momentum of either a photon or a particle is,
$$p=\frac{h}{\lambda}$$
Where ‘h’ is Plank’s constant.
Q.1.21 If the particles listed below all have same wavelength, which has shortest wavelength; electron, alpha-particle, neutron and proton.
Answer: From de-Broglie relation for wavelength in terms of kinetic energy
$$\lambda=\frac{h}{\sqrt{2m_0K}}$$
We see that the particle which is heaviest, will have shortest wavelength. As mass of alpha-particle is greatest of all particles listed, so it will have shortest wavelength.
Q.1.22 Does a photon have a de-Broglie wavelength? Explain.
Answer: In de-Broglie hypothesis an electromagnetic wave is the de-Broglie wave for a photon moving with speed ‘c’. The de-Broglie waves for electron, proton, neutron etc. are not electromagnetic waves but matter waves. The wavelength of photon according to de-Broglie hypothesis is;
$$\lambda=\frac{h}{P}$$
Where ‘p’ is momentum of photon.
Q.1.23 Name at least two experiments which support the wave nature of matter.
Answer: Following two experiments support wave nature of matter:
- Davison and Germer experiment
- G.P. Thomson experiment
Q.1.24 Is an electron a particle? Is it a wave? Explain your answer, citing relevant experimental evidence.
Answer: we observe the wave property of electron in a diffraction experiment like Davison-particle property in J.J. Thomson’s measurement of for electron etc. but not the wave property. It is not possible to perform a single experiment in which electron will behave as a particle as well as wave.
Q.1.25 Name at least two experiments which support the particle nature of radiation.
Answer: Following two experiments support particle nature of radiation:
- Photoelectric effect
- Compton Effect
Q.1.26 Name at least one experiments in each case which support the wave nature of matter and wave nature of radiation.
Answer: G.P Thomson experiment supports wave nature of matter and Young double slit interference experiment supports wave nature of radiation.
Q.1.27 Considering electron and photon as particles how are they different from each other?
Answer: 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 a photon with frequency ʋ is $P=\frac{hv}{c}$
Electron:
- An electron has mass and may be at rest.
- An electron cannot move with velocity of light.
- Momentum of electron with mass ‘m’ moving with velocity v is, p=mv
Q.1.28 If an electron and proton have same de-Broglie wavelength, which particle has greater speed?
Answer: According to de-Broglie, the wavelength associated with a particle is given by,
$$\lambda=\frac{h}{mv}$$
$$\Rightarrow v=\frac{h}{m\lambda}$$
For same wavelength, the smaller mass has greater velocity. So electron will have greater speed.
Q.1.29 A proton has a slightly smaller mass than a neutron. Would a proton of same wavelength have more or less kinetic energy?
Answer: From de-Broglie relation for wavelength in terms of kinetic energy
$$\lambda=\frac{h}{\sqrt{2m_0K}}\ \ \ \ \ \ \ \Rightarrow\lambda^2=\frac{h^2}{2m_0K}\ \ \ \ \ \ \ \ \ \ \Rightarrow K=\frac{h^2}{2m_0\lambda^2}$$
We see that for a given wavelength, kinetic energy is inversely proportional to the mass. Hence proton with smaller mass has more kinetic energy than neutron.
