Ans: 1
Potentialof the electrons, V= 30 kV = 3 ×104 V
Hence, energy of the electrons, E = 3 ×104 eV
Where,
e= Charge on an electron = 1.6 ×10−19C
(a)Maximum frequency produced by the X-rays = ν
The energy of the electrons is given by the relation:
E= hν
Where,
h= Planck’s constant = 6.626 ×10−34Js
Hence, the maximum frequency of X-rays produced is
(b)The minimum wavelength produced by the X-rays is given as:
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
Ans: 2
Work function of caesium metal,
Frequency of light,
(a)The maximum kinetic energy is given by the photoelectric effect as:
Where,
h= Planck’s constant = 6.626 ×10−34Js
Hence, the maximum kinetic energy of the emitted electrons is
0.345 eV.
(b)For stopping potential
, we can write the equation for kinetic energy as:
Hence, the stopping potential of the material is 0.345 V.
(c)Maximum speed of the emitted photoelectrons = v
Hence,the relation for kinetic energy can be written as:
Where,
m= Mass of an electron = 9.1 ×10−31kg
Hence, the maximum speed of the emitted photoelectrons is
332.3 km/s.
Ans: 3
Photoelectric cut-off voltage, V0= 1.5 V
Themaximum kinetic energy of the emitted photoelectrons is given as:
Where,
e= Charge on an electron = 1.6 ×10−19C
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 ×10−19J.
Ans: 4
Wavelength of the monochromatic light, λ= 632.8 nm = 632.8 ×10−9m
Power emitted by the laser, P= 9.42 mW = 9.42 ×10−3W
Planck’s constant, h= 6.626 ×10−34Js
Speed of light, c= 3 ×108m/s
Mass of a hydrogen atom, m= 1.66 ×10−27kg
(a)The energy of each photon is given as:
The momentum of each photon is given as:
(b)Number of photons arriving per second, at a target irradiated by the beam =n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence,the equation for power can be written as:
(c)Momentum of the hydrogen atom is the same as the momentum of the photon,
Momentum is given as:
Where,
v= Speed of the hydrogen atom
Ans: 5
Energy flux of sunlight reaching the surface of earth, Φ= 1.388 ×103W/m2
Hence, power of sunlight per square metre, P= 1.388 ×103 W
Speed of light, c= 3 ×108m/s
Planck’s constant, h= 6.626 ×10−34Js
Average wavelength of photons present in sunlight,
Number of photons per square metre incident on earth per second = n
Hence,the equation for power can be written as:
Therefore,every second,
photons are incident per square metre on earth.
Ans: 6
The slope of the cut-off voltage (V) versus frequency (ν)of an incident light is given as:
Where,
e= Charge on an electron = 1.6 ×10−19C
h= Planck’s constant
Therefore,the value of Planck’s constant is
Ans: 7
Power of the sodium lamp, P= 100 W
Wavelength of the emitted sodium light, λ= 589 nm = 589 ×10−9m
Planck’s constant, h= 6.626 ×10−34 Js
Speed of light, c = 3 ×108m/s
(a)The energy per photon associated with the sodium light is given as:
(b)Number of photons delivered to the sphere = n
The equation for power can be written as:
Therefore, every second,
photons are delivered to the sphere.
Ans: 8
Threshold frequency of the metal,
Frequencyof light incident on the metal,
Charge on an electron, e= 1.6 ×10−19C
Planck’s constant, h= 6.626 ×10−34Js
Cut-off voltage for the photoelectric emission from the metal =
The equation for the cut-off energy is given as:
Therefore, the cut-off voltage for the photoelectric emission is
Ans: 9
No
Work function of the metal,
Charge on an electron, e= 1.6 ×10−19C
Planck’sconstant, h= 6.626 ×10−34Js
Wavelength of the incident radiation, λ= 330 nm = 330 × 10−9m
Speed of light, c= 3 ×108m/s
The energy of the incident photon is given as:
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.
Ans: 10
Frequency of the incident photon,
Maximum speed of the electrons, v= 6.0 ×105m/s
Planck’s constant, h= 6.626 ×10−34Js
Mass of an electron, m= 9.1 ×10−31kg
For threshold frequency ν0, the relation for kinetic energy is written as:
Therefore, the threshold frequency for the photoemission of electrons is
Ans: 11
Wavelength of light produced by the argon laser, λ= 488 nm
= 488 × 10−9m
Stopping potential of the photoelectrons, V0= 0.38 V
1eV = 1.6 × 10−19J
∴ V0=
Planck’s constant, h= 6.6 × 10−34Js
Charge on an electron, e= 1.6 × 10−19C
Speed of light, c= 3 × 10 m/s
From Einstein’s photoelectric effect, we have the relation involving the work function Φ0of the material of the emitter as:
Therefore, the material with which the emitter is made has the work function of 2.16 eV.
Ans: 12
Potential difference, V= 56 V
Planck’s constant, h= 6.6 × 10−34Js
Mass of an electron, m= 9.1 × 10−31kg
Charge on an electron, e= 1.6 × 10−19C
(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:
The momentum of each accelerated electron is given as:
p= mv
= 9.1 × 10−31× 4.44 × 106
= 4.04 × 10−24kg m s−1
Therefore, the momentum of each electron is 4.04 × 10−24kg m s−1.
(b) De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
Ans: 13
Kinetic energy of the electron, Ek= 120 eV
Planck’s constant, h= 6.6 × 10−34Js
Mass of an electron, m= 9.1 × 10−31kg
Charge on an electron, e= 1.6 × 10−19C
(a) For the electron, we can 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 × 10−24kg m s−1
Therefore, the momentum of the electron is 5.91 × 10−24kg m s−1.
(b) Speed of the electron, v= 6.496 × 106m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as:
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
Ans: 14
Wavelength of light of a sodium line, λ= 589 nm = 589 × 10−9m
Mass of an electron, me= 9.1 × 10−31kg
Mass of a neutron, mn= 1.66 × 10−27kg
Planck’s constant, h= 6.6 × 10−34Js
(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
We have the relation for de Broglie wavelength as:
Substituting equation (2) in equation (1), we get the relation:
Hence, the kinetic energy of the electron is 6.9 × 10−25J or 4.31 μeV.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
Hence, the kinetic energy of the neutron is 3.78 × 10−28J or 2.36 neV.
Ans: 15
(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 × 10−34Js
De Broglie wavelength of the bullet is given by the relation:
(b) Mass of the ball, m= 0.060 kg
Speed of the ball, v= 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
(c)Mass of the dust particle, m= 1 × 10−9kg
Speed of the dust particle, v= 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
Ans: 16
Wavelength of an electron
= 1 × 10−9m
Planck’s constant, h= 6.63 × 10−34Js
(a) The momentum of an elementary particle is given by de Broglie relation:
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.
(b) The energy of a photon is given by the relation:
Where,
Speed of light, c= 3 × 108m/s
Therefore, the energy of the 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 × 10−31kg
p= 6.63 × 10−25kg m s−1
Hence, the kinetic energy of the electron is 1.51 eV.
Ans: 17
(a) De Broglie wavelength of the neutron, λ= 1.40 × 10−10m
Mass of a neutron, mn= 1.66 × 10−27kg
Planck’s constant, h= 6.6 × 10−34Js
Kinetic energy (K) and velocity (v) are related as:
…(1)
De Broglie wavelength (λ)and velocity (v) are related as:
Using equation (2) in equation (1), we get:
Hence, the kinetic energy of the neutron is 6.75 × 10−21J or 4.219 × 10−2eV.
(b) Temperature of the neutron, T= 300 K
Boltzmann constant, k= 1.38 × 10−23kg m2s−2K−1
Average kinetic energy of the neutron:
The relation for the de Broglie wavelength is given as:
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
Ans: 18
The momentum of a photon having energy (hν)is given as:
Where,
λ =Wavelength of the electromagnetic radiation
c= Speed of light
h= Planck’s constant
De Broglie wavelength of the 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 the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
Ans: 19
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 × 10−27kg
∴m= 28.0152 ×1.66 × 10−27kg
Planck’s constant, h= 6.63 × 10−34Js
Boltzmann constant, k= 1.38 × 10−23J K−1
We have the expression that relates mean kinetic energy
of the nitrogen molecule with the root mean square speed 
as:
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
Ans: 20
(a)Potential difference across the evacuated tube, V= 500 V
Specific charge of an electron, e/m= 1.76 ×1011 C kg−1
The speed of each emitted electron is given by the relation for kinetic energy as:
Therefore, the speed of each emitted electron is
(b)Potential of the anode, V= 10 MV = 10 ×106 V
The speed of each electron is given as:
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., forv <<c.
For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:
E= mc2
Where,
m= Relativistic mass
m0= Mass of the particle at rest
Kinetic energy is given as:
K= mc2− m0c2
Ans: 21
(a)Speed of an electron, v= 5.20 ×106m/s
Magnetic field experienced by the electron, B= 1.30 ×10−4T
Specific charge of an electron, e/m= 1.76 ×1011 C kg−1
Where,
e= Charge on the electron = 1.6 ×10−19C
m= Mass of the electron = 9.1 ×10−31kg−1
The force exerted on the electron is given as:
θ= Angle between the magnetic field and the beam velocity
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.
Hence, equation (1) reduces to:
Therefore, the radius of the circular path is 22.7 cm.
(b) Energy of the electron beam, E = 20 MeV
The energy of the electron is given as:
This result is incorrect 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., forv <<c
When very high speeds are concerned, the relativistic domain comes into consideration.
In the relativistic domain, mass is given as:
Where,
= Mass of the particle at rest
Hence, the radius of the circular path is given as:
Ans: 22
Potential of an anode, V = 100 V
Magnetic field experienced by the electrons, B= 2.83 ×10−4T
Radius of the circular orbit r= 12.0 cm = 12.0 × 10−2m
Mass of each electron = m
Charge on each electron = e
Velocity of each electron = v
The energy of each electron is equal to its kinetic energy, i.e.,
It is the magnetic field, due to its bending nature, that provides the centripetal force
for the beam. Hence, we can write:
Centripetal force = Magnetic force
Putting the value of vin equation (1), we get:
Therefore, the specific charge ratio (e/m) is
Ans: 23
(a) Wavelength produced by an X-ray tube,
Planck’s constant, h= 6.626 ×10−34Js
Speed of light, c = 3 ×108m/s
The maximum energy of a photon is given as:
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.
Ans: 24
Total energy of two γ-rays:
E= 10. 2 BeV
= 10.2 × 109eV
= 10.2 ×109 × 1.6 ×10−10J
Hence, the energy of each γ-ray:
Planck’s constant,
Speed of light,
Energy is related to wavelength as:
Therefore, the wavelength associated with each γ-ray is
Ans: 25
(a) Power of the medium wave transmitter, P= 10 kW = 104 W = 104J/s
Hence, energy emitted by the transmitter per second, E= 104
Wavelength of the radio wave, λ= 500 m
The energy of the wave is given as:
Where,
h= Planck’s constant = 6.6 × 10−34Js
c= Speed of light = 3 × 108m/s
Letn be the number of photons emitted by the transmitter.
∴nE1= E
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.
(b) Intensity of light perceived by the human eye, I= 10−10W m−2
Area of a pupil, A= 0.4 cm2 = 0.4 × 10−4m2
Frequency of white light, ν=6 × 1014Hz
The energy emitted by a photon is given as:
E= hν
Where,
h= Planck’s constant = 6.6 × 10−34Js
∴E= 6.6 × 10−34× 6 × 1014
= 3.96 × 10−19J
Letn be the total number of photons falling per second, per unit area of the pupil.
The total energy per unit for nfalling photons is given as:
E= n × 3.96 × 10−19J s−1m−2
The energy per unit area per second is the intensity of light.
∴E = I
n× 3.96 × 10−19= 10−10
= 2.52 × 108m2 s−1
The total number of photons entering the pupil per second is given as:
nA= n × A
= 2.52 × 108× 0.4 × 10−4
= 1.008 × 104s−1
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.
Ans: 26
Wavelength of ultraviolet light, λ= 2271 Å = 2271 × 10−10m
Stopping potential of the metal, V0= 1.3 V
Planck’s constant, h= 6.6 × 10−34J
Charge on an electron, e= 1.6 × 10−19C
Work function of the metal =
Frequency of light = ν
We have the photo-energy relation from the photoelectric effect as:
= hν− eV0
Letν0be the threshold frequency of the metal.
∴=hν0
Wavelength of red light,
= 6328 × 10−10m
∴Frequency of red light,
Sinceν0> νr, the photocell will not respond to the red light produced by the laser.
Ans: 27
Wavelength of the monochromatic radiation, λ= 640.2 nm
= 640.2 × 10−9m
Stopping potential of the neon lamp, V0= 0.54 V
Charge on an electron, e= 1.6 × 10−19C
Planck’s constant, h= 6.6 × 10−34Js
Let
be the work function and νbe the frequency of emitted light.
We have the photo-energy relation from the photoelectric effect as:
eV0= hν−
Wavelength of the radiation emitted from an iron source, λ' = 427.2 nm
=427.2 × 10−9m
Let
be the new stopping potential. Hence, photo-energy is given as:
Hence, the new stopping potential is 1.50 eV.
Ans: 28
Einstein’s photoelectric equation is given as:
eV0= hν−
Where,
V0= Stopping potential
h= Planck’s constant
e= Charge on an electron
ν =Frequency of radiation
= Work function of a material
It can be concluded from equation (1) that potential V0is directly proportional to frequency ν.
Frequency is also given by the relation:
This relation can be used to obtain the frequencies of the various lines of the given wavelengths.
The given quantities can be listed in tabular form as:
It can be observed that the obtained curve is a straight line. It intersects the ν-axis at 5 × 1014Hz, which is the threshold frequency (ν0) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the λ5line, and therefore, no stopping voltage is required to stop the current.
Slope of the straight line =
From equation (1), the slope
can be written as:
The work function of the metal is given as:
=hν0
= 6.573 × 10−34× 5 × 1014
= 3.286 × 10−19J
= 2.054 eV
Ans: 29
Mo and Ni will not show photoelectric emission in both cases
Wavelength for a radiation, λ= 3300 Å = 3300 × 10−10m
Speed of light, c= 3 × 108m/s
Planck’s constant, h= 6.6 × 10−34 Js
The energy of incident radiation is given as:
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.
Ans: 30
Intensity of incident light, I= 10−5W m−2
Surface area of a sodium photocell, A= 2 cm2= 2 × 10−4m2
Incident power of the light, P = I × A
= 10−5× 2 × 10−4
= 2 × 10−9W
Work function of the metal,
= 2 eV
= 2 × 1.6 × 10−19
= 3.2 × 10−19J
Number of layers of sodium that absorbs the incident energy, n= 5
We know that the effective atomic area of a sodium atom, Aeis 10−20m2.
Hence, the number of conduction electrons in nlayers is given as:
The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:
Time required for photoelectric emission:
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.
Ans: 31
An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ= 1 Å = 10−10m
Mass of an electron, me= 9.11 × 10−31kg
Planck’s constant, h= 6.6 × 10−34Js
Charge on an electron, e= 1.6 × 10−19C
The kinetic energy of the electron is given as:
Where,
v= Velocity of the electron
mev= Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
Energy of a photon,
Hence, a photon has a greater energy than an electron for the same wavelength.
Ans: 32
(a) De Broglie wavelength =
; neutron is not suitable for the diffraction experiment
Kinetic energy of the neutron, K= 150 eV
= 150 × 1.6 × 10−19
= 2.4 × 10−17J
Mass of a neutron, mn= 1.675 × 10−27kg
The kinetic energy of the neutron is given by the relation:
Where,
v= Velocity of the neutron
mnv= Momentum of the neutron
De-Broglie wavelength of the neutron is given as:
It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e., 10−10m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy
150 eV is not suitable for diffraction experiments.
(b) De Broglie wavelength =
Room temperature, T= 27°C = 27 + 273 = 300 K
The average kinetic energy of the neutron is given as:
Where,
k= Boltzmann constant = 1.38 × 10−23J mol−1K−1
The wavelength of the neutron is given as:
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.
Ans: 33
Electrons are accelerated by a voltage, V= 50 kV = 50 × 103V
Charge on an electron, e= 1.6 × 10−19C
Mass of an electron, me= 9.11 × 10−31kg
Wavelength of yellow light = 5.9 × 10−7m
The kinetic energy of the electron is given as:
E= eV
= 1.6 × 10−19× 50 × 103
= 8 × 10−15J
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 105times that of an optical microscope.
Ans: 34
Wavelength of a proton or a neutron, λ≈10−15m
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 × 10−34Js
Speed of light, c= 3 × 108m/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.
Ans: 35
De Broglie wavelength associated with He atom =
Room temperature, T= 27°C = 27 + 273 = 300 K
Atmospheric pressure, P= 1 atm = 1.01 × 105Pa
Atomic weight of a He atom = 4
Avogadro’s number, NA= 6.023 × 1023
Boltzmann constant, k= 1.38 × 10−23J mol−1K−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.
Ans: 36
Temperature,T = 27°C = 27 + 273 = 300 K
Mean separation between two electrons, r= 2 × 10−10m
De Broglie wavelength of an electron is given as:
Where,
h= Planck’s constant = 6.6 × 10−34Js
m= Mass of an electron = 9.11 × 10−31kg
k= Boltzmann constant = 1.38 × 10−23J mol−1K−1
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
Ans: 37
(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 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 (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.
Potentialof the electrons, V= 30 kV = 3 ×104 V
Hence, energy of the electrons, E = 3 ×104 eV
Where,
e= Charge on an electron = 1.6 ×10−19C
(a)Maximum frequency produced by the X-rays = ν
The energy of the electrons is given by the relation:
E= hν
Where,
h= Planck’s constant = 6.626 ×10−34Js
Hence, the maximum frequency of X-rays produced is
(b)The minimum wavelength produced by the X-rays is given as:
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
Ans: 2
Work function of caesium metal,
Frequency of light,
(a)The maximum kinetic energy is given by the photoelectric effect as:
Where,
h= Planck’s constant = 6.626 ×10−34Js
Hence, the maximum kinetic energy of the emitted electrons is
0.345 eV.
(b)For stopping potential
, we can write the equation for kinetic energy as:Hence, the stopping potential of the material is 0.345 V.
(c)Maximum speed of the emitted photoelectrons = v
Hence,the relation for kinetic energy can be written as:
Where,
m= Mass of an electron = 9.1 ×10−31kg
Hence, the maximum speed of the emitted photoelectrons is
332.3 km/s.
Ans: 3
Photoelectric cut-off voltage, V0= 1.5 V
Themaximum kinetic energy of the emitted photoelectrons is given as:
Where,
e= Charge on an electron = 1.6 ×10−19C
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 ×10−19J.
Ans: 4
Wavelength of the monochromatic light, λ= 632.8 nm = 632.8 ×10−9m
Power emitted by the laser, P= 9.42 mW = 9.42 ×10−3W
Planck’s constant, h= 6.626 ×10−34Js
Speed of light, c= 3 ×108m/s
Mass of a hydrogen atom, m= 1.66 ×10−27kg
(a)The energy of each photon is given as:
The momentum of each photon is given as:
(b)Number of photons arriving per second, at a target irradiated by the beam =n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence,the equation for power can be written as:
(c)Momentum of the hydrogen atom is the same as the momentum of the photon,
Momentum is given as:
Where,
v= Speed of the hydrogen atom
Ans: 5
Energy flux of sunlight reaching the surface of earth, Φ= 1.388 ×103W/m2
Hence, power of sunlight per square metre, P= 1.388 ×103 W
Speed of light, c= 3 ×108m/s
Planck’s constant, h= 6.626 ×10−34Js
Average wavelength of photons present in sunlight,
Number of photons per square metre incident on earth per second = n
Hence,the equation for power can be written as:
Therefore,every second,
photons are incident per square metre on earth.Ans: 6
The slope of the cut-off voltage (V) versus frequency (ν)of an incident light is given as:
Where,
e= Charge on an electron = 1.6 ×10−19C
h= Planck’s constant
Therefore,the value of Planck’s constant is
Ans: 7
Power of the sodium lamp, P= 100 W
Wavelength of the emitted sodium light, λ= 589 nm = 589 ×10−9m
Planck’s constant, h= 6.626 ×10−34 Js
Speed of light, c = 3 ×108m/s
(a)The energy per photon associated with the sodium light is given as:
(b)Number of photons delivered to the sphere = n
The equation for power can be written as:
Therefore, every second,
photons are delivered to the sphere.Ans: 8
Threshold frequency of the metal,
Frequencyof light incident on the metal,
Charge on an electron, e= 1.6 ×10−19C
Planck’s constant, h= 6.626 ×10−34Js
Cut-off voltage for the photoelectric emission from the metal =
The equation for the cut-off energy is given as:
Therefore, the cut-off voltage for the photoelectric emission is
Ans: 9
No
Work function of the metal,
Charge on an electron, e= 1.6 ×10−19C
Planck’sconstant, h= 6.626 ×10−34Js
Wavelength of the incident radiation, λ= 330 nm = 330 × 10−9m
Speed of light, c= 3 ×108m/s
The energy of the incident photon is given as:
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.
Ans: 10
Frequency of the incident photon,
Maximum speed of the electrons, v= 6.0 ×105m/s
Planck’s constant, h= 6.626 ×10−34Js
Mass of an electron, m= 9.1 ×10−31kg
For threshold frequency ν0, the relation for kinetic energy is written as:
Therefore, the threshold frequency for the photoemission of electrons is
Ans: 11
Wavelength of light produced by the argon laser, λ= 488 nm
= 488 × 10−9m
Stopping potential of the photoelectrons, V0= 0.38 V
1eV = 1.6 × 10−19J
∴ V0=
Planck’s constant, h= 6.6 × 10−34Js
Charge on an electron, e= 1.6 × 10−19C
Speed of light, c= 3 × 10 m/s
From Einstein’s photoelectric effect, we have the relation involving the work function Φ0of the material of the emitter as:
Therefore, the material with which the emitter is made has the work function of 2.16 eV.
Ans: 12
Potential difference, V= 56 V
Planck’s constant, h= 6.6 × 10−34Js
Mass of an electron, m= 9.1 × 10−31kg
Charge on an electron, e= 1.6 × 10−19C
(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:
The momentum of each accelerated electron is given as:
p= mv
= 9.1 × 10−31× 4.44 × 106
= 4.04 × 10−24kg m s−1
Therefore, the momentum of each electron is 4.04 × 10−24kg m s−1.
(b) De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
Ans: 13
Kinetic energy of the electron, Ek= 120 eV
Planck’s constant, h= 6.6 × 10−34Js
Mass of an electron, m= 9.1 × 10−31kg
Charge on an electron, e= 1.6 × 10−19C
(a) For the electron, we can 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 × 10−24kg m s−1
Therefore, the momentum of the electron is 5.91 × 10−24kg m s−1.
(b) Speed of the electron, v= 6.496 × 106m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as:
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
Ans: 14
Wavelength of light of a sodium line, λ= 589 nm = 589 × 10−9m
Mass of an electron, me= 9.1 × 10−31kg
Mass of a neutron, mn= 1.66 × 10−27kg
Planck’s constant, h= 6.6 × 10−34Js
(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
We have the relation for de Broglie wavelength as:
Substituting equation (2) in equation (1), we get the relation:
Hence, the kinetic energy of the electron is 6.9 × 10−25J or 4.31 μeV.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
Hence, the kinetic energy of the neutron is 3.78 × 10−28J or 2.36 neV.
Ans: 15
(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 × 10−34Js
De Broglie wavelength of the bullet is given by the relation:
(b) Mass of the ball, m= 0.060 kg
Speed of the ball, v= 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
(c)Mass of the dust particle, m= 1 × 10−9kg
Speed of the dust particle, v= 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
Ans: 16
Wavelength of an electron
= 1 × 10−9m
Planck’s constant, h= 6.63 × 10−34Js
(a) The momentum of an elementary particle is given by de Broglie relation:
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.
(b) The energy of a photon is given by the relation:
Where,
Speed of light, c= 3 × 108m/s
Therefore, the energy of the 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 × 10−31kg
p= 6.63 × 10−25kg m s−1
Hence, the kinetic energy of the electron is 1.51 eV.
Ans: 17
(a) De Broglie wavelength of the neutron, λ= 1.40 × 10−10m
Mass of a neutron, mn= 1.66 × 10−27kg
Planck’s constant, h= 6.6 × 10−34Js
Kinetic energy (K) and velocity (v) are related as:
…(1)De Broglie wavelength (λ)and velocity (v) are related as:
Using equation (2) in equation (1), we get:
Hence, the kinetic energy of the neutron is 6.75 × 10−21J or 4.219 × 10−2eV.
(b) Temperature of the neutron, T= 300 K
Boltzmann constant, k= 1.38 × 10−23kg m2s−2K−1
Average kinetic energy of the neutron:
The relation for the de Broglie wavelength is given as:
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
Ans: 18
The momentum of a photon having energy (hν)is given as:
Where,
λ =Wavelength of the electromagnetic radiation
c= Speed of light
h= Planck’s constant
De Broglie wavelength of the 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 the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
Ans: 19
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 × 10−27kg
∴m= 28.0152 ×1.66 × 10−27kg
Planck’s constant, h= 6.63 × 10−34Js
Boltzmann constant, k= 1.38 × 10−23J K−1
We have the expression that relates mean kinetic energy
of the nitrogen molecule with the root mean square speed as:Hence, the de Broglie wavelength of the nitrogen molecule is given as:
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
Ans: 20
(a)Potential difference across the evacuated tube, V= 500 V
Specific charge of an electron, e/m= 1.76 ×1011 C kg−1
The speed of each emitted electron is given by the relation for kinetic energy as:
Therefore, the speed of each emitted electron is
(b)Potential of the anode, V= 10 MV = 10 ×106 V
The speed of each electron is given as:
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., forv <<c.
For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:
E= mc2
Where,
m= Relativistic mass
m0= Mass of the particle at rest
Kinetic energy is given as:
K= mc2− m0c2
Ans: 21
(a)Speed of an electron, v= 5.20 ×106m/s
Magnetic field experienced by the electron, B= 1.30 ×10−4T
Specific charge of an electron, e/m= 1.76 ×1011 C kg−1
Where,
e= Charge on the electron = 1.6 ×10−19C
m= Mass of the electron = 9.1 ×10−31kg−1
The force exerted on the electron is given as:
θ= Angle between the magnetic field and the beam velocity
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.Hence, equation (1) reduces to:
Therefore, the radius of the circular path is 22.7 cm.
(b) Energy of the electron beam, E = 20 MeV
The energy of the electron is given as:
This result is incorrect 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., forv <<c
When very high speeds are concerned, the relativistic domain comes into consideration.
In the relativistic domain, mass is given as:
Where,
= Mass of the particle at restHence, the radius of the circular path is given as:
Ans: 22
Potential of an anode, V = 100 V
Magnetic field experienced by the electrons, B= 2.83 ×10−4T
Radius of the circular orbit r= 12.0 cm = 12.0 × 10−2m
Mass of each electron = m
Charge on each electron = e
Velocity of each electron = v
The energy of each electron is equal to its kinetic energy, i.e.,
It is the magnetic field, due to its bending nature, that provides the centripetal force
for the beam. Hence, we can write:Centripetal force = Magnetic force
Putting the value of vin equation (1), we get:
Therefore, the specific charge ratio (e/m) is
Ans: 23
(a) Wavelength produced by an X-ray tube,
Planck’s constant, h= 6.626 ×10−34Js
Speed of light, c = 3 ×108m/s
The maximum energy of a photon is given as:
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.
Ans: 24
Total energy of two γ-rays:
E= 10. 2 BeV
= 10.2 × 109eV
= 10.2 ×109 × 1.6 ×10−10J
Hence, the energy of each γ-ray:
Planck’s constant,
Speed of light,
Energy is related to wavelength as:
Therefore, the wavelength associated with each γ-ray is
Ans: 25
(a) Power of the medium wave transmitter, P= 10 kW = 104 W = 104J/s
Hence, energy emitted by the transmitter per second, E= 104
Wavelength of the radio wave, λ= 500 m
The energy of the wave is given as:
Where,
h= Planck’s constant = 6.6 × 10−34Js
c= Speed of light = 3 × 108m/s
Letn be the number of photons emitted by the transmitter.
∴nE1= E
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.
(b) Intensity of light perceived by the human eye, I= 10−10W m−2
Area of a pupil, A= 0.4 cm2 = 0.4 × 10−4m2
Frequency of white light, ν=6 × 1014Hz
The energy emitted by a photon is given as:
E= hν
Where,
h= Planck’s constant = 6.6 × 10−34Js
∴E= 6.6 × 10−34× 6 × 1014
= 3.96 × 10−19J
Letn be the total number of photons falling per second, per unit area of the pupil.
The total energy per unit for nfalling photons is given as:
E= n × 3.96 × 10−19J s−1m−2
The energy per unit area per second is the intensity of light.
∴E = I
n× 3.96 × 10−19= 10−10
= 2.52 × 108m2 s−1
The total number of photons entering the pupil per second is given as:
nA= n × A
= 2.52 × 108× 0.4 × 10−4
= 1.008 × 104s−1
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.
Ans: 26
Wavelength of ultraviolet light, λ= 2271 Å = 2271 × 10−10m
Stopping potential of the metal, V0= 1.3 V
Planck’s constant, h= 6.6 × 10−34J
Charge on an electron, e= 1.6 × 10−19C
Work function of the metal =
Frequency of light = ν
We have the photo-energy relation from the photoelectric effect as:
= hν− eV0Letν0be the threshold frequency of the metal.
∴
=hν0Wavelength of red light,
= 6328 × 10−10m∴Frequency of red light,
Sinceν0> νr, the photocell will not respond to the red light produced by the laser.
Ans: 27
Wavelength of the monochromatic radiation, λ= 640.2 nm
= 640.2 × 10−9m
Stopping potential of the neon lamp, V0= 0.54 V
Charge on an electron, e= 1.6 × 10−19C
Planck’s constant, h= 6.6 × 10−34Js
Let
be the work function and νbe the frequency of emitted light.We have the photo-energy relation from the photoelectric effect as:
eV0= hν−
Wavelength of the radiation emitted from an iron source, λ' = 427.2 nm
=427.2 × 10−9m
Let
be the new stopping potential. Hence, photo-energy is given as:Hence, the new stopping potential is 1.50 eV.
Ans: 28
Einstein’s photoelectric equation is given as:
eV0= hν−
Where,
V0= Stopping potential
h= Planck’s constant
e= Charge on an electron
ν =Frequency of radiation
= Work function of a materialIt can be concluded from equation (1) that potential V0is directly proportional to frequency ν.
Frequency is also given by the relation:
This relation can be used to obtain the frequencies of the various lines of the given wavelengths.
The given quantities can be listed in tabular form as:
Frequency × 1014 Hz 8.2197.4126.8845.4934.343Stopping potential V0 1.280.950.740.160
Slope of the straight line =
From equation (1), the slope
can be written as:The work function of the metal is given as:
=hν0= 6.573 × 10−34× 5 × 1014
= 3.286 × 10−19J
= 2.054 eVAns: 29
Mo and Ni will not show photoelectric emission in both cases
Wavelength for a radiation, λ= 3300 Å = 3300 × 10−10m
Speed of light, c= 3 × 108m/s
Planck’s constant, h= 6.6 × 10−34 Js
The energy of incident radiation is given as:
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.
Ans: 30
Intensity of incident light, I= 10−5W m−2
Surface area of a sodium photocell, A= 2 cm2= 2 × 10−4m2
Incident power of the light, P = I × A
= 10−5× 2 × 10−4
= 2 × 10−9W
Work function of the metal,
= 2 eV= 2 × 1.6 × 10−19
= 3.2 × 10−19J
Number of layers of sodium that absorbs the incident energy, n= 5
We know that the effective atomic area of a sodium atom, Aeis 10−20m2.
Hence, the number of conduction electrons in nlayers is given as:
The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:
Time required for photoelectric emission:
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.
Ans: 31
An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ= 1 Å = 10−10m
Mass of an electron, me= 9.11 × 10−31kg
Planck’s constant, h= 6.6 × 10−34Js
Charge on an electron, e= 1.6 × 10−19C
The kinetic energy of the electron is given as:
Where,
v= Velocity of the electron
mev= Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
Energy of a photon,
Hence, a photon has a greater energy than an electron for the same wavelength.
Ans: 32
(a) De Broglie wavelength =
; neutron is not suitable for the diffraction experimentKinetic energy of the neutron, K= 150 eV
= 150 × 1.6 × 10−19
= 2.4 × 10−17J
Mass of a neutron, mn= 1.675 × 10−27kg
The kinetic energy of the neutron is given by the relation:
Where,
v= Velocity of the neutron
mnv= Momentum of the neutron
De-Broglie wavelength of the neutron is given as:
It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e., 10−10m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy
150 eV is not suitable for diffraction experiments.
(b) De Broglie wavelength =
Room temperature, T= 27°C = 27 + 273 = 300 K
The average kinetic energy of the neutron is given as:
Where,
k= Boltzmann constant = 1.38 × 10−23J mol−1K−1
The wavelength of the neutron is given as:
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.
Ans: 33
Electrons are accelerated by a voltage, V= 50 kV = 50 × 103V
Charge on an electron, e= 1.6 × 10−19C
Mass of an electron, me= 9.11 × 10−31kg
Wavelength of yellow light = 5.9 × 10−7m
The kinetic energy of the electron is given as:
E= eV
= 1.6 × 10−19× 50 × 103
= 8 × 10−15J
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 105times that of an optical microscope.
Ans: 34
Wavelength of a proton or a neutron, λ≈10−15m
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 × 10−34Js
Speed of light, c= 3 × 108m/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.
Ans: 35
De Broglie wavelength associated with He atom =
Room temperature, T= 27°C = 27 + 273 = 300 K
Atmospheric pressure, P= 1 atm = 1.01 × 105Pa
Atomic weight of a He atom = 4
Avogadro’s number, NA= 6.023 × 1023
Boltzmann constant, k= 1.38 × 10−23J mol−1K−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.
Ans: 36
Temperature,T = 27°C = 27 + 273 = 300 K
Mean separation between two electrons, r= 2 × 10−10m
De Broglie wavelength of an electron is given as:
Where,
h= Planck’s constant = 6.6 × 10−34Js
m= Mass of an electron = 9.11 × 10−31kg
k= Boltzmann constant = 1.38 × 10−23J mol−1K−1
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
Ans: 37
(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
andIt 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 (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.
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