Class 12 Physics Study Guide: CBSE Class 12 Physics Dual Nature Of Matter and Radiation chapter notes and important questions
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Class 12 Physics: Dual Nature of Matter and Radiation
ELECTRON EMISSION
- There are three types of electron emission, namely thermionic emission, photoelectric emission and field emission.
- The minimum energy required by an electron to escape from the metal surface is called the work function.
- Work function is conveniently expressed in electron volts (eV).
- One electron volt is the energy gained or lost by an electron while passing through a potential difference of one volt.
PHOTOELECTRIC EFFECT
- The minimum energy required by an electron to come out from a metal surface is called the work function of the metal.
- The photoelectric effect is the phenomenon of emission of electrons from metals when illuminated by light of suitable frequency.
- Photoelectric current depends on:
- the intensity of the incident light
- the potential difference applied between the two electrodes
- the nature of the emitter material
EXPERIMENTAL STUDY OF PHOTOELECTRIC EFFECT
Variation of photoelectric current with collector plate potential for different frequencies of incident radiation (\(\nu_3 > \nu_2 > \nu_1\)).
Variation of stopping potential \(V_0\) with frequency of the incident radiation for two metals A and B.
- The minimum negative potential given to the anode plate for which the photoelectric current becomes zero is called the stopping potential.
- The stopping potential \(V_0\) depends on (i) the frequency of the incident light and (ii) the nature of the emitter material. For a given frequency of incident light, the stopping potential is independent of its intensity.
- Below a certain frequency (the threshold frequency \(\nu_0\), characteristic of the metal), no photoelectric emission takes place, no matter how large the intensity may be.
\[eV_0 = \tfrac12 m v_{max}^2 = K_{max}\]
EINSTEIN'S PHOTOELECTRIC EQUATION: ENERGY QUANTUM OF RADIATION
- Light is composed of discrete packets of energy called quanta or photons.
- The energy carried by each photon is \(E = h\nu\), where \(\nu\) is the frequency, and its momentum is \(p = h/\lambda\). The energy of the photon depends on the frequency \(\nu\) of the incident light and not on its intensity.
- Photoelectric emission from the metal surface occurs due to absorption of a photon by an electron.
- Einstein's photoelectric equation: \(K_{max} = h\nu - \phi_0\) or \(eV_0 = h\nu - \phi_0\).
PARTICLE NATURE OF LIGHT: THE PHOTON
- Radiation has dual nature: wave and particle. The wave nature is revealed in phenomena like interference, diffraction and polarisation. The particle nature is revealed by the photoelectric effect.
- By symmetry, matter also should have dual nature: wave and particle. The waves associated with a moving material particle are called matter waves or de Broglie waves.
- The de Broglie wavelength \(\lambda\) associated with a moving particle is related to its momentum \(p\) as \(\lambda = h/p = h/mv\).
- The de Broglie wavelength of an electron accelerated through a potential \(V\): consider an electron of mass \(m\) and charge \(e\) accelerated from rest through a potential \(V\).
\[K = eV,\qquad K = \tfrac12 mv^2 = \frac{p^2}{2m}\]
\[p^2 = 2mK\ \Rightarrow\ p = \sqrt{2mK} = \sqrt{2meV}\]
\[\lambda = \frac{h}{\sqrt{2meV}}\]
Substituting the numerical values of \(h\), \(m\) and \(e\):
\[\lambda = \frac{12.27}{\sqrt V}\ \text{\AA}\]
QUESTION BANK
ONE MARK QUESTIONS
A proton and an electron have the same kinetic energy. Which one has the greater de Broglie wavelength and why?
Answer: The electron. For the same kinetic energy \(K\), \(\lambda = \dfrac{h}{\sqrt{2mK}}\), so \(\lambda \propto \dfrac1{\sqrt m}\). The electron has the smaller mass, so its wavelength is greater.
Define the term 'stopping potential' in relation to the photoelectric effect.
Answer: It is the minimum negative (retarding) potential given to the collector plate with respect to the emitter at which the photoelectric current becomes zero.
The stopping potential in an experiment on the photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?
Answer: \(K_{max} = eV_0 = 1.5\) eV (\(= 2.4\times10^{-19}\) J).
The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?
Answer: \(V_0 = K_{max}/e = 3\) V.
With what purpose was the famous Davisson-Germer experiment with electrons performed?
Answer: It was performed to verify the wave nature of electrons (de Broglie's hypothesis) by observing the diffraction of an electron beam from a nickel crystal.
If the intensity of incident radiation in a photoelectric experiment is doubled, what happens to the kinetic energy of the emitted photoelectrons?
Answer: It remains unchanged. The kinetic energy depends only on the frequency of the radiation and the work function; doubling the intensity only doubles the number of photoelectrons emitted per second.
Calculate the frequency associated with a photon of energy \(3.3\times10^{-10}\) J.
Answer: \(\nu = E/h = \dfrac{3.3\times10^{-10}}{6.63\times10^{-34}} \approx 5\times10^{23}\) Hz.
What is the momentum of a photon of energy 1 MeV?
Answer: \(E = 1\) MeV \(= 1.6\times10^{-13}\) J, so \(p = E/c = \dfrac{1.6\times10^{-13}}{3\times10^{8}} = 5.33\times10^{-22}\) kg m/s.
What happens to the velocity of emitted electrons when the wavelength of the incident light is decreased?
Answer: It increases. A smaller wavelength means a higher frequency and a more energetic photon, so the maximum kinetic energy \(K_{max} = \dfrac{hc}{\lambda} - \phi_0\) increases.
Which of the two photons is more energetic: red light or violet light?
Answer: Violet light, because it has the higher frequency (\(E = h\nu\)).
What will be the stopping potential when a photon of 25 eV is incident on a metal surface of work function 6 eV?
Answer: \(eV_0 = 25 - 6 = 19\) eV, so \(V_0 = 19\) volt.
Why is an alkali metal surface better suited as a photosensitive surface?
Answer: Alkali metals have a low work function, so even visible light has enough energy to eject electrons from them.
Blue light can eject electrons from a photosensitive surface while orange light cannot. Will violet and red light eject electrons from the same surface?
Answer: Violet light will, because its frequency is higher than that of blue light (which is above the threshold). Red light will not, because its frequency is lower than that of orange light (which is below the threshold).
Matter waves associated with electrons could be verified by crystal diffraction experiments. Why?
Answer: The wavelength of the matter waves associated with electrons is comparable to the spacing between the atomic planes of crystals.
How do matter waves differ from light waves as regards the velocity of the particle and the wave?
Answer: In the case of matter waves, the wave velocity is different from the particle velocity. But in the case of light, the particle (photon) velocity and the wave velocity are the same.
An electron and an alpha particle have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?
Answer: The alpha particle.
An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has the largest de Broglie wavelength?
Answer: The electron. For the same kinetic energy, \(\lambda \propto 1/\sqrt m\), and the electron has the smallest mass.
In an experiment on the photoelectric effect, the following graphs were obtained between the photoelectric current (I) and the anode potential (V). Name the characteristic of the incident radiation that was kept constant in this experiment.
Answer: The frequency of the incident radiation was kept constant. All the curves have the same stopping potential \(-V_0\), which depends only on frequency; the different saturation currents correspond to different intensities.
Write the expression for the de Broglie wavelength associated with a charged particle having charge 'q' and mass 'm' when it is accelerated by a potential V.
Answer: \(\lambda = \dfrac{h}{\sqrt{2mqV}}\)
TWO MARKS QUESTIONS
How is the photoelectric current affected on increasing the (i) frequency, (ii) intensity of the incident radiation, and why?
Answer: (i) On increasing the frequency (above the threshold), the photoelectric current does not change, because the current depends on the number of photoelectrons emitted per second, i.e. on the number of photons per second, not on the energy of each photon. Only the kinetic energy of the electrons increases. (ii) On increasing the intensity, the photoelectric current increases in proportion, because more photons strike the surface per second and so more electrons are emitted per second.
The graph shows the variation of stopping potential \(V_0\) versus frequency \(\nu\) of incident radiation for two photosensitive metals A and B. Which of the two metals has the higher threshold frequency and why?
Answer: Metal B. The threshold frequency is the frequency at which the line cuts the frequency axis (\(V_0 = 0\)). The line for B cuts the axis at \(\nu_0'\), which is greater than \(\nu_0\) for A.
The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has the higher value of work function? Justify your answer.
Answer: Metal B. The work function is \(\phi_0 = h\nu_0\). Since B has the higher threshold frequency (\(\nu_0' > \nu_0\)), its work function is higher. The intercept on the \(V_0\) axis, \(-\phi_0/e\), is also larger in magnitude for B.
An alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths.
Answer: \(\lambda = \dfrac{h}{\sqrt{2mqV}}\), so \(\dfrac{\lambda_\alpha}{\lambda_p} = \sqrt{\dfrac{m_pq_p}{m_\alpha q_\alpha}} = \sqrt{\dfrac{1\times1}{4\times2}} = \dfrac{1}{2\sqrt2}\). That is, \(\lambda_p : \lambda_\alpha = 2\sqrt2 : 1\).
Name the element with the highest work function and also the element with the lowest work function.
Answer: Highest work function: platinum (5.65 eV). Lowest work function: caesium (2.14 eV).
Calculate the work function of a metal in eV if its threshold wavelength is 6800 Å.
Answer: \(\phi_0 = \dfrac{hc}{\lambda_0} = \dfrac{12400\ \text{eV}\,\text{\AA}}{6800\ \text{\AA}} \approx 1.82\) eV.
Work function of aluminium is 4.2 eV. If two photons, each of energy 2.5 eV, are incident on its surface, will emission of electrons take place?
Answer: No. In the photoelectric effect one electron absorbs only one photon at a time. Each photon has only 2.5 eV, which is less than the work function 4.2 eV, and the energies of two photons cannot be added.
A radio transmitter operates at a frequency of 880 kHz and a power of 10 kW. Find the number of photons emitted per second.
Answer: Energy of one photon \(= h\nu = 6.63\times10^{-34}\times880\times10^{3} = 5.83\times10^{-28}\) J. Number of photons per second \(n = \dfrac{\text{energy emitted per second}}{\text{energy of one photon}} = \dfrac{10^{4}}{5.83\times10^{-28}} = 1.716\times10^{31}\).
A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface.
Answer: Force \(= \dfrac{P}{c}\left(0.4\times1 + 0.6\times2\right) = \dfrac{10}{3\times10^{8}}\times1.6 = 5.33\times10^{-8}\) N.
Find the ratio of the wavelength of a 10 keV photon to that of a 10 keV electron.
Answer: \(\lambda_{photon} = \dfrac{hc}{E} = 1.24\ \text{\AA}\) and \(\lambda_{electron} = \dfrac{12.27}{\sqrt{10^4}} = 0.1227\ \text{\AA}\). Ratio \(= \dfrac{1.24}{0.1227} \approx 10 : 1\).
A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of the wavelengths associated with the two.
Answer: Using \(\lambda = h/\sqrt{2mqV}\): \(\lambda_p : \lambda_\alpha = 2\sqrt2 : 1\).
Why do macroscopic objects in our daily life not show wave-like properties? OR Why is the wave nature of particles significant in the sub-atomic domain only?
Answer: Macroscopic objects in our daily life do not show wave-like properties because the wavelength associated with them is very small and beyond the scope of any measurement. In the sub-atomic world, the masses of the particles are extremely small, leading to a wavelength that is measurable.
Show that Bohr's second postulate, 'the electron revolves around the nucleus only in certain fixed orbits without radiating energy', can be explained on the basis of de Broglie's hypothesis of the wave nature of the electron.
Answer: An electron orbit can be stable only if the electron wave forms a standing wave, i.e. the circumference of the orbit contains a whole number of de Broglie wavelengths: \(2\pi r = n\lambda\). With \(\lambda = h/mv\) this gives \(mvr = \dfrac{nh}{2\pi}\), which is Bohr's second postulate. Since only these orbits give complete standing waves, the non-radiating electron can exist only in certain fixed orbits.
The de Broglie wavelength associated with an electron accelerated through a potential difference V is \(\lambda\). What will be the de Broglie wavelength when the accelerating potential difference is increased to 4V?
Answer: \(\lambda \propto \dfrac{1}{\sqrt V}\), so \(\dfrac{\lambda_2}{\lambda} = \sqrt{\dfrac{V}{4V}} = \dfrac12\). The new wavelength is \(\lambda_2 = \dfrac{\lambda}{2}\).
Determine the accelerating potential required for an electron to have a de Broglie wavelength of 1 Å.
Answer: \(\lambda = \dfrac{12.27}{\sqrt V}\ \text{\AA} = 1\ \text{\AA}\), so \(V = (12.27)^2 \approx 150.6\) V.
An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has (i) the shortest and (ii) the largest de Broglie wavelength?
Answer: \(\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt m}\). (i) The alpha particle (largest mass) has the shortest wavelength. (ii) The electron (smallest mass) has the largest wavelength.
In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be \(4.13\times10^{-15}\) V s. Calculate the value of Planck's constant.
Answer: From \(V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}\), slope \(= h/e\). So \(h = 4.13\times10^{-15}\times1.6\times10^{-19} = 6.6\times10^{-34}\) J s.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer: Photon energy \(= \dfrac{1240\ \text{eV nm}}{330\ \text{nm}} = 3.76\) eV. This is less than the work function 4.2 eV, so there will be no photoelectric emission.
THREE MARKS QUESTIONS
An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of the electron.
Answer: (a) Both have the same momentum: \(p = \dfrac h\lambda = \dfrac{6.63\times10^{-34}}{10^{-9}} = 6.63\times10^{-25}\) kg m/s. (b) Photon energy \(E = pc = 6.63\times10^{-25}\times3\times10^{8} = 1.99\times10^{-16}\) J \(\approx 1.24\) keV. (c) Electron kinetic energy \(K = \dfrac{p^2}{2m} = \dfrac{(6.63\times10^{-25})^2}{2\times9.1\times10^{-31}} = 2.4\times10^{-19}\) J \(\approx 1.5\) eV.
Calculate the (a) momentum and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer: (a) \(p = \sqrt{2meV} = \sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times56} = 4.04\times10^{-24}\) kg m/s. (b) \(\lambda = \dfrac hp = \dfrac{6.63\times10^{-34}}{4.04\times10^{-24}} = 1.64\times10^{-10}\) m \(= 0.164\) nm.
The work function of caesium metal is 2.14 eV. When light of frequency \(6\times10^{14}\) Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential, and (c) maximum speed of the emitted photoelectrons?
Answer: Photon energy \(h\nu = 6.63\times10^{-34}\times6\times10^{14} = 3.98\times10^{-19}\) J \(= 2.49\) eV. (a) \(K_{max} = 2.49 - 2.14 = 0.35\) eV \(\approx 5.6\times10^{-20}\) J. (b) \(V_0 = K_{max}/e = 0.35\) V. (c) \(v_{max} = \sqrt{\dfrac{2K_{max}}{m}} = \sqrt{\dfrac{2\times5.6\times10^{-20}}{9.1\times10^{-31}}} \approx 3.5\times10^{5}\) m/s.
Set up Einstein's photoelectric equation using the photon picture of electromagnetic radiation. Explain briefly how this equation accounts for all the observations in the photoelectric effect.
Answer: In the photon picture, light consists of photons of energy \(h\nu\). One photon is absorbed by one electron. Part of this energy, \(\phi_0\) (the work function), is used to free the electron from the metal and the rest appears as its maximum kinetic energy: \(K_{max} = h\nu - \phi_0\), i.e. \(eV_0 = h\nu - \phi_0\). How it explains the observations: (1) Emission needs \(h\nu \ge \phi_0\), so there is a threshold frequency \(\nu_0 = \phi_0/h\) below which no emission takes place, whatever the intensity. (2) \(K_{max}\) depends only on \(\nu\) and \(\phi_0\), not on intensity, and \(V_0\) against \(\nu\) is a straight line of slope \(h/e\). (3) Intensity is the number of photons per second, so the number of photoelectrons (the photocurrent) is proportional to intensity. (4) Each electron absorbs a single photon, so emission is instantaneous.
In a plot of photoelectric current versus anode potential, how does (i) the saturation current vary with anode potential for incident radiations of different frequencies but the same intensity? (ii) The stopping potential vary for incident radiations of different intensities but the same frequency? (iii) The photoelectric current vary for different intensities but the same frequency of incident radiations? Justify your answer in each case.
Answer: (i) The saturation current does not depend on the anode potential (all emitted electrons are already collected), and for the same intensity it is the same for the different frequencies, because it depends on the number of photons striking per second, not on their frequency. (ii) The stopping potential is the same for different intensities of the same frequency, because \(eV_0 = h\nu - \phi_0\) depends only on the frequency. (iii) The photoelectric current (and the saturation current) increases in proportion to the intensity, because a larger intensity means more photons per second and hence more photoelectrons per second.
Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, \(\nu_1 > \nu_2\), of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer.
Answer: The stopping potential is higher for \(\nu_1\). Since \(eV_0 = h\nu - \phi_0\), a higher frequency gives a larger \(K_{max}\) and therefore a larger stopping potential (\(V_{01} > V_{02}\)). The saturation currents are the same as the intensity is the same.
Define threshold frequency in connection with the photoelectric effect. The two lines A and B shown in the graph plot the de Broglie wavelength \(\lambda\) as a function of \(1/\sqrt V\) (V is the accelerating potential) for two particles having the same charge. Which of the two represents the particle of heavier mass?
Answer: Threshold frequency is the minimum frequency of the incident radiation below which no photoelectric emission takes place, however intense the radiation is. The slope of the graph is \(h/\sqrt{2mq}\) (for the same charge). A smaller slope means a larger mass, so the line with the smaller slope (A) represents the heavier particle.
Draw a graph between kinetic energy and frequency and find (a) the threshold frequency, (b) the work function from the intercept, (c) Planck's constant from the slope of the graph.
Answer: The graph of \(K_{max}\) against \(\nu\) is a straight line, as \(K_{max} = h\nu - \phi_0\). (a) The intercept on the frequency axis is the threshold frequency \(\nu_0\). (b) The intercept on the energy axis is \(-\phi_0\), so the work function is its magnitude. (c) The slope of the line is Planck's constant \(h\).
4. Einstein’s photo electric equation:
K =1/2 mv2= 2mK
P = √2mK = √2meV
λ = h/√2meV
Ans: ν = 5 x 10 23 Hz.
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