Page No 411: - Chapter 11 Dual Nature Of Radiation & Matter Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]
Updated By SaraNextGen
On April 24, 2024, 11:35 AM
Question 11.33:
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?
Answer:
Electrons are accelerated by a voltage, V = 50 kV = 50 × 103 V
Charge on an electron, e = 1.6 × 10−19 C
Mass of an electron, me = 9.11 × 10−31 kg
Wavelength of yellow light = 5.9 × 10−7 m
The kinetic energy of the electron is given as:
E = eV
= 1.6 × 10−19 × 50 × 103
= 8 × 10−15 J
De Broglie wavelength is given by the relation:
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.
Question 11.34:
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10−15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)
Answer:
Wavelength of a proton or a neutron, λ ≈ 10−15 m
Rest mass energy of an electron:
m0c2 = 0.511 MeV
= 0.511 × 106 × 1.6 × 10−19
= 0.8176 × 10−13 J
Planck’s constant, h = 6.6 × 10−34 Js
Speed of light, c = 3 × 108 m/s
The momentum of a proton or a neutron is given as:
The relativistic relation for energy (E) is given as:
Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.
Question 11.35:
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.
Answer:
De Broglie wavelength associated with He atom =
Room temperature, T = 27°C = 27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 × 105 Pa
Atomic weight of a He atom = 4
Avogadro’s number, NA = 6.023 × 1023
Boltzmann constant, k = 1.38 × 10−23 J mol−1 K−1
Average energy of a gas at temperature T,is given as:
De Broglie wavelength is given by the relation:
Where,
m = Mass of a He atom
We have the ideal gas formula:
PV = RT
PV = kNT
Where,
V = Volume of the gas
N = Number of moles of the gas
Mean separation between two atoms of the gas is given by the relation:
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.
Question 11.36:
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 × 10−10 m.
[Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]
Answer:
Temperature, T = 27°C = 27 + 273 = 300 K
Mean separation between two electrons, r = 2 × 10−10 m
De Broglie wavelength of an electron is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
m = Mass of an electron = 9.11 × 10−31 kg
k = Boltzmann constant = 1.38 × 10−23 J mol−1 K−1
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
Question 11.37:
Answer the following Questions:
(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?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
E = hν, p = 
But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed νλ) has no physical significance. Why?
Answer:
(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.
(b) The basic relations for electric field and magnetic field are
.
These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius), and B (magnetic field). These relations give the value of velocity of an electron as
and
It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, 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 (ν) associated with an electron has no direct physical significance.
Therefore, the product νλ(phase speed)has no physical significance.
Group speed is given as:
This quantity has a physical meaning.
Also Read : INTRODUCTION-Chapter-12-Atoms-Exercise-Solutions-class-12-ncert-solutions-Physics
