Chapter 11 Dual Nature Of Radiation And Matter Solutions
Question - 31 : - Anelectron microscope uses electrons accelerated by a voltage of 50 kV. Determinethe de Broglie wavelength associated with the electrons. If other factors (suchas numerical aperture, etc.) are taken to be roughly the same, how does theresolving power of an electron microscope compare with that of an opticalmicroscope which uses yellow light?
Answer - 31 : -
Electrons are accelerated by avoltage, 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 theelectron is given as:
E = eV
= 1.6 × 10−19 × 50 × 103
= 8 × 10−15 J
De Broglie wavelength isgiven by the relation:
This wavelength is nearly 105 timesless than the wavelength of yellow light.
Theresolving power of a microscope is inversely proportional to the wavelength oflight used. Thus, the resolving power of an electron microscope is nearly 105 timesthat of an optical microscope.
Question - 32 : - The wavelength of a probe is roughly ameasure of the size of a structure that it can probe in some detail. The quark structureof protons and neutrons appears at the minute length-scale of 10−15 mor less. This structure was first probed in early 1970’s using high energyelectron beams produced by a linear accelerator at Stanford, USA. Guess whatmight have been the order of energy of these electron beams. (Rest mass energyof electron = 0.511 MeV.)
Answer - 32 : -
Wavelength of a proton or a neutron, λ ≈10−15 m
Rest mass energy of anelectron:
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 protonor 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 ofthe order of 1.24 BeV.
Question - 33 : - Findthe typical de Broglie wavelength associated with a He atom in helium gas atroom temperature (27 ºC) and 1 atm pressure; and compare it with the meanseparation between two atoms under these conditions.
Answer - 33 : - De Broglie wavelength associated with He atom =
Room temperature, T = 27°C= 27 + 273 = 300 K
Atmospheric pressure, P = 1atm = 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,isgiven as:
De Broglie wavelength isgiven by the relation:
Where,
m = Mass of a He atom
We have the ideal gasformula:
PV = RT
PV = kNT
Where,
V = Volume of the gas
N = Number of moles of the gas
Mean separation betweentwo atoms of the gas is given by the relation:
Hence,the mean separation between the atoms is much greater than the de Brogliewavelength.