Chapter 11 Dual Nature Of Radiation And Matter Solutions
Question - 11 : - What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength ofan electron with kinetic energy of 120 eV.
Answer - 11 : -
Kinetic energy of the electron, Ek =120 eV
Planck’s constant, h = 6.6× 10−34 Js
Mass of an electron, m =9.1 × 10−31 kg
Charge on an electron, e =1.6 × 10−19 C
(a) For the electron, wecan write the relation for kinetic energy as:
Where,
v = Speed of the electron
Momentum of the electron, p = mv
= 9.1 × 10−31 × 6.496 × 106
= 5.91 × 10−24 kg m s−1
Therefore, the momentum of the electron is5.91 × 10−24 kg m s−1.
(b) Speed of theelectron, v = 6.496 × 106 m/s
(c) De Brogliewavelength of an electron having a momentum p, is given as:
Therefore,the de Broglie wavelength of the electron is 0.112 nm.
Question - 12 : - The wavelength of lightfrom the spectral emission line of sodium is 589 nm. Find the kinetic energy atwhich
(a) an electron, and
(b) a neutron, wouldhave the same de Broglie wavelength.
Answer - 12 : -
Wavelength of light of a sodium line, λ =589 nm = 589 × 10−9 m
Mass of an electron, me= 9.1 ×10−31 kg
Mass of a neutron, mn= 1.66 × 10−27 kg
Planck’s constant, h = 6.6× 10−34 Js
(a) For the kinetic energy K, ofan electron accelerating with a velocity v, we have the relation:We have the relation forde Broglie wavelength as:
Substituting equation (2)in equation (1), we get the relation:
Hence, the kinetic energy of the electron is6.9 × 10−25 J or 4.31 μeV.
(b) Using equation (3), we canwrite the relation for the kinetic energy of the neutron as:
Hence, the kinetic energy of the electron is6.9 × 10−25 J or 4.31 μeV.
Hence,the kinetic energy of the neutron is 3.78 × 10−28 J or 2.36neV.
What is the de Brogliewavelength of
(a) a bullet of mass 0.040 kgtravelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kgmoving at a speed of 1.0 m/s, and
(c) a dust particle of mass1.0 × 10−9 kg drifting with a speed of 2.2 m/s?
Answer - 13 : -
(a)Mass of the bullet, m =0.040 kg
Speed of the bullet, v =1.0 km/s = 1000 m/s
Planck’s constant, h = 6.6× 10−34 Js
DeBroglie wavelength of the bullet is given by the relation:(b) Mass of the ball, m =0.060 kg
Speed of the ball, v = 1.0m/s
De Broglie wavelength ofthe ball is given by the relation:
(c)Mass of the dustparticle, m = 1 × 10−9 kg
Speed of the dust particle, v =2.2 m/s
De Broglie wavelength ofthe dust particle is given by the relation:
An electron and a photoneach have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of thephoton, and
(c) the kinetic energy of electron.
Answer - 14 : - Wavelength of an electron 
= 1 × 10−9 m
Planck’s constant, h = 6.63× 10−34 Js
(a) The momentum of anelementary particle is given by de Broglie relation:
It is clear that momentumdepends only on the wavelength of the particle. Since the wavelengths of anelectron and a photon are equal, both have an equal momentum.
(b) The energy of aphoton is given by the relation:
Where,
Speed of light, c = 3 × 108 m/s
Therefore, the energy ofthe photon is 1.243 keV.
(c) The kinetic energy (K)of an electron having momentum p,is given by the relation:
Where,
m = Mass of the electron = 9.1 × 10−31 kg
p = 6.63 × 10−25 kg m s−1
Hence,the kinetic energy of the electron is 1.51 eV.
(a) For what kinetic energy ofa neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?
(b) Also find the deBroglie wavelength of a neutron, in thermal equilibrium with matter, having anaverage kinetic energy of (3/2) kT at 300 K.
Answer - 15 : -
(a) De Brogliewavelength of the neutron, λ = 1.40 × 10−10 m
Mass of a neutron, mn =1.66 × 10−27 kg
Planck’s constant, h = 6.6× 10−34 Js
Kineticenergy (K) and velocity (v) are related as: … (1)
DeBroglie wavelength (λ) and velocity (v) are related as:Using equation (2) inequation (1), we get:
Hence, the kinetic energy of the neutron is6.75 × 10−21 J or 4.219 × 10−2 eV.
(b) Temperature of theneutron, T = 300 K
Boltzmann constant, k =1.38 × 10−23 kg m2 s−2 K−1
Average kinetic energy ofthe neutron:
The relation for the deBroglie wavelength is given as:
Therefore,the de Broglie wavelength of the neutron is 0.146 nm.
Showthat the wavelength of electromagnetic radiation is equal to the de Brogliewavelength of its quantum (photon).
Answer - 16 : -
The momentum of a photon having energy (hν)is given as:
Where,
λ = Wavelength of theelectromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength ofthe photon is given as:
Where,
m = Mass of the photon
v = Velocity of the photon
Hence,it can be inferred from equations (i) and (ii) that thewavelength of the electromagnetic radiation is equal to the de Brogliewavelength of the photon.
Question - 17 : - Whatis the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assumethat the molecule is moving with the root-mean square speed of molecules atthis temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer - 17 : -
Temperature of the nitrogen molecule, T =300 K
Atomic mass of nitrogen =14.0076 u
Hence, mass of the nitrogen molecule, m =2 × 14.0076 = 28.0152 u
But 1 u = 1.66 × 10−27 kg
∴m = 28.0152 ×1.66 × 10−27 kg
Planck’s constant, h = 6.63× 10−34 Js
Boltzmann constant, k =1.38 × 10−23 J K−1
Wehave the expression that relates mean kinetic energy 
of the nitrogen molecule with the rootmean square speed 
as:
Hence, the de Brogliewavelength of the nitrogen molecule is given as:
Therefore,the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
(a) Estimate the speed with whichelectrons emitted from a heated emitter of an evacuated tube impinge on thecollector maintained at a potential difference of 500 V with respect to theemitter. Ignore the small initial speeds of the electrons. The specificcharge of the electron, i.e., its e/m is given to be1.76 × 1011 C kg−1.
(b) Use the same formula you employ in (a) to obtainelectron speed for an collector potential of 10 MV. Do you see what is wrong?In what way is the formula to be modified?
Answer - 18 : -
(a)Potential difference acrossthe evacuated tube, V = 500 V
Specific charge of an electron, e/m =1.76 × 1011 C kg−1
The speed of each emittedelectron is given by the relation for kinetic energy as:
Therefore, the speed of each emitted electronis 
(b)Potential of the anode, V =10 MV = 10 × 106 V
Thespeed of each electron is given as:
This result is wrong because nothing canmove 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.
For very high speedproblems, relativistic equations must be considered for solving them. In therelativistic limit, the total energy is given as:
E = mc2
Where,
m =Relativistic mass
m0 = Mass of the particle at rest
Kinetic energy is givenas:
K = mc2 − m0c2
Question - 19 : - (a) A monoenergeticelectron beam with electron speed of 5.20 × 106 m s−1 issubject to a magnetic field of 1.30 × 10−4 T normal to the beamvelocity. What is the radius of the circle traced by the beam, given e/m forelectron equals 1.76 × 1011 C kg−1.
(b) Is the formula youemploy in (a) valid for calculating radius of the path of a 20 MeV electronbeam? If not, in what way is it modified?
[Note: Exercises11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond thescope of this book. They have been inserted here simply to emphasise the pointthat the formulas you use in part (a) of the exercises are not valid at veryhigh speeds or energies. See answers at the end to know what ‘very high speedor energy’ means.]
Answer - 19 : -
(a)Speed of anelectron, v = 5.20 × 106 m/s
Magnetic field experienced by theelectron, B = 1.30 × 10−4 T
Specific charge of an electron, e/m =1.76 × 1011 C kg−1
Where,
e = Charge on the electron = 1.6 × 10−19 C
m = Mass of the electron = 9.1 × 10−31 kg−1
The force exerted on theelectron is given as:
θ = Angle between themagnetic field and the beam velocity
Themagnetic 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 thecentripetal force
for the beam.
Hence, equation (1)reduces to:
Therefore, the radius ofthe circular path is 22.7 cm.
(b) Energy of the electron beam, E =20 MeV
The energy of the electronis given as:
This result is incorrect because nothing canmove 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
When very high speeds areconcerned, the relativistic domain comes into consideration.
In the relativisticdomain, mass is given as:
Where,
= Mass of the particle atrest
Hence, the radius of thecircular path is given as:
Question - 20 : - An electron gun with its collector at apotential of 100 V fires out electrons in a spherical bulb containing hydrogengas at low pressure (∼10−2 mm of Hg). A magneticfield of 2.83 × 10−4 T curves the path of the electrons in acircular orbit of radius 12.0 cm. (The path can be viewed because the gas ionsin the path focus the beam by attracting electrons, and emitting light byelectron capture; this method is known as the ‘fine beam tube’ method.Determine e/m from the data.
Answer - 20 : -
Potential of an anode, V =100 V
Magnetic field experienced by the electrons, B =2.83 × 10−4 T
Radius of the circular orbit r =12.0 cm = 12.0 × 10−2 m
Mass of each electron = m
Charge on each electron = e
Velocity of each electron = v
The energy of eachelectron is equal to its kinetic energy, i.e.,
It is the magnetic field, due to its bendingnature, that provides the centripetal force for the beam. Hence,we can write:
Centripetal force =Magnetic force
Putting the value of v inequation (1), we get:
Therefore, the specific charge ratio (e/m)is