Source: https://www.toppersbulletin.com/ncert-solutions-for-class-12-physics-chapter-11-dual-nature-of-radiation-and-matter/
Timestamp: 2019-04-25 06:10:35+00:00

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(b) minimum wavelength of X-rays produced by 30 kV electrons.
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
(c) maximum speed of the emitted photoelectrons?
Hence, the stopping potential of the material is 0.345 V.
Q3 :The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 ×10-19J.
Q4 :Monochromatic light of wavelength 632.8 nm is produced by ahelium-neon laser. The power emitted is 9.42 mW.
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Assume that the beam has a uniform cross-section that is less than the target area.
Q5 :The energy flux of sunlight reaching the surface of the earth is 1.388 x 103W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Therefore, every second,3.84 X 1021 photons are incident per square metre on earth.
Q6 :In an experiment on photoelectric effect, the slope of the cut-offvoltage versus frequency of incident light is found to be 4.12 x 10-15V s. Calculate the value of Planck’s constant.
Q7 :A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Therefore,2.96 x 1020 every second, photons are delivered to the sphere.
Q8 :The threshold frequency for a certain metal is 3.3 x 1014 Hz. If light of frequency 8.2 x 1014Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Q9 :The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.
Q10 :Light of frequency 7.21 x 1014Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 105m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Q11 :Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Therefore, the material with which the emitter is made has the work function of 2.16 eV.
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Therefore, the momentum of each electron is 4.04 × 10-24kg m s-1.
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Therefore, the momentum of the electron is 5.91 × 10-24kg m s-1.
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
(b) a neutron, would have the same de Broglie wavelength.
Hence, the kinetic energy of the electron is 6.9 × 10-25J or 4.31 μeV.
Hence, the kinetic energy of the neutron is 3.78 × 10-28J or 2.36 neV.
(c) a dust particle of mass 1.0 x 10-9kg drifting with a speed of 2.2 m/s?
(c) the kinetic energy of electron.
It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.
Therefore, the energy of the photon is 1.243 keV.
Hence, the kinetic energy of the electron is 1.51 eV.
Q17 :(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 x 10-10 m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.
Hence, the kinetic energy of the neutron is 6.75 × 10-21J or 4.219 × 10-2eV.
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
Q18 :Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
Q20 :(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 x 1011C kg-1.
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
This result is wrong because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << c.
s-1is subject to a magnetic field of 1.30 x 10-4T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 x 1011C kg-1.
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
The magnetic field is normal to the direction of beam.
The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force for the beam.
Therefore, the radius of the circular path is 22.7 cm.
When very high speeds are concerned, the relativistic domain comes into consideration.
(∼10-2mm of Hg). A magnetic field of 2.83 x 10-4T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method. Determine e/mfrom the data.
Q23 :(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Therefore, the maximum energy of an X-ray photon is 27.6 keV.
(b) Accelerating voltage provides energy to the electrons for producing X-rays. To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.
Q25 :Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10-10W m-2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 x 1014Hz.
Let n be the number of photons emitted by the transmitter.
The energy (E1) of a radio photon is very less, but the number of photons (n) emitted per second in a radio wave is very large.
The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.
Let n be the total number of photons falling per second, per unit area of the pupil.
The energy per unit area per second is the intensity of light.
This number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.
Q26 :Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼105W m-2) red light of wavelength 6328 Å produced by a He-Ne laser?
Let V0 be the threshold frequency of the metal.
Since V0> Vr, the photocell will not respond to the red light produced by the laser.
Q27 :Monochromatic radiation of wavelength 640.2 nm (1nm = 10-9m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Let be the work function and V be the frequency of emitted light.
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
It can be concluded from equation (1) that potential V0 is directly proportional to frequency V.
This relation can be used to obtain the frequencies of the various lines of the given wavelengths.
The following figure shows a graph between V and V0.
It can be observed that the obtained curve is a straight line. It intersects the V-axis at 5 × 1014Hz, which is the threshold frequency (V0) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the λ5 line, and therefore, no stopping voltage is required to stop the current.
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
It can be observed that the energy of the incident radiation is greater than the work function of Na and K only. It is less for Mo and Ni. Hence, Mo and Ni will not show photoelectric emission.
If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase. This does not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.
Q30 :Light of intensity 10-5W m-2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
We know that the effective atomic area of a sodium atom, Ae is 10-20m2.
The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.
Q31 :Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me= 9.11 x 10-31kg).
An X-ray probe has a greater energy than an electron probe for the same wavelength.
Hence, a photon has a greater energy than an electron for the same wavelength.
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 ºC). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
150 eV is not suitable for diffraction experiments.
This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high-energy neutron beam should first be thermalised, before using it for diffraction.
Q33 :An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
This wavelength is nearly 105 times less than the wavelength of yellow light.
The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly 105 times that of an optical microscope.
Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.
Q35 :Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.
Q36 :Compute the typical de Broglie wavelength of an electron in a metal at 27 ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10-10 m.
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; ( – 1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
But while the value of λ is physically significant, the value of V(and therefore, the value of the phase speed Vλ) has no physical significance. Why?
(a) Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.
It can be observed from these relations that the dynamics of an electron is determined not by eand mseparately, but by the ratio e/m.
(c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.
(d) The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.
(e) The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength (λ) is significant, but the frequency (V) associated with an electron has no direct physical significance.
Therefore, the product Vλ (phase speed)has no physical significance.
This quantity has a physical meaning.

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