Chapter 11 Dual Nature Of Radiation And Matter Solutions
Question - 21 : - (a) An X-ray tubeproduces a continuous spectrum of radiation with its short wavelength end at0.45 Å. What is the maximum energy of a photon in the radiation?
(b) Fromyour answer to (a), guess what order of accelerating voltage (for electrons) isrequired in such a tube?
Answer - 21 : - (a) Wavelengthproduced by an X-ray tube, 
Planck’s constant, h =6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
The maximum energy of aphoton is given as:
Therefore, the maximumenergy of an X-ray photon is 27.6 keV.
(b) Accelerating voltage provides energy to theelectrons for producing X-rays. To get an X-ray of 27.6 keV, the incidentelectrons must possess at least 27.6 keV of kinetic electric energy. Hence, anaccelerating voltage of the order of 30 keV is required for producing X-rays.
In an accelerator experiment on high-energycollisions of electrons with positrons, a certain event is interpreted asannihilation of an electron-positron pair of total energy 10.2 BeV into twoγ-rays of equal energy. What is the wavelength associated with each γ-ray?(1BeV = 109 eV)
Answer - 22 : -
Total energy of two γ-rays:
E = 10. 2 BeV
= 10.2 × 109 eV
= 10.2 × 109 × 1.6 × 10−10 J
Hence, the energy of each γ-ray:
Planck’s constant, 
Speed of light, 
Energy is related to wavelengthas:
Therefore, the wavelength associated witheach γ-ray is
Question - 23 : - Estimating the followingtwo numbers should be interesting. The first number will tell you why radioengineers do not need to worry much about photons! The second number tells youwhy our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photonsemitted per second by a Medium wave transmitter of 10 kW power, emittingradiowaves of wavelength 500 m.
(b) The number of photonsentering the pupil of our eye per second corresponding to the minimum intensityof white light that we humans can perceive (∼10−10 W m−2).Take the area of the pupil to be about 0.4 cm2, and the averagefrequency of white light to be about 6 × 1014 Hz.
Answer - 23 : -
(a) Power of the mediumwave transmitter, P = 10 kW = 104 W = 104 J/s
Hence, energy emitted by the transmitter persecond, E = 104
Wavelength of the radio wave, λ =500 m
Theenergy of the wave is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
Let n be the number ofphotons emitted by the transmitter.
∴nE1 = E
The energy (E1) of a radiophoton is very less, but the number of photons (n) emitted per second ina radio wave is very large.
The existence of a minimumquantum of energy can be ignored and the total energy of a radio wave can betreated as being continuous.
(b) Intensity of lightperceived by the human eye, I = 10−10 W m−2
Area of a pupil, A = 0.4 cm2 =0.4 × 10−4 m2
Frequency of white light, ν= 6 ×1014 Hz
The energy emitted by aphoton is given as:
E = hν
Where,
h = Planck’s constant = 6.6 × 10−34 Js
∴E = 6.6 × 10−34 ×6 × 1014
= 3.96 × 10−19 J
Let n be the total numberof photons falling per second, per unit area of the pupil.
The total energy per unit for n fallingphotons is given as:
E = n × 3.96 × 10−19 Js−1 m−2
The energy per unit areaper second is the intensity of light.
∴E = I
n ×3.96 × 10−19 = 10−10
= 2.52 × 108 m2 s−1
The total number ofphotons entering the pupil per second is given as:
nA = n × A
= 2.52 × 108 × 0.4 × 10−4
= 1.008 × 104 s−1
Thisnumber is not as large as the one found in problem (a), but it is large enough for the human eye to never see theindividual photons.
Ultraviolet light of wavelength 2271 Å froma 100 W mercury source irradiates a photo-cell made of molybdenum metal. If thestopping potential is −1.3 V, estimate the work function of the metal. Howwould the photo-cell respond to a high intensity (∼105 W m−2) redlight of wavelength 6328 Å produced by a He-Ne laser?
Answer - 24 : -
Wavelength of ultraviolet light, λ =2271 Å = 2271 × 10−10 m
Stopping potential of the metal, V0 =1.3 V
Planck’s constant, h = 6.6× 10−34 J
Charge on an electron, e =1.6 × 10−19 C
Workfunction of the metal = 
Frequency of light = ν
We have the photo-energyrelation from the photoelectric effect as:
= hν − eV0
Let ν0 be thethreshold frequency of the metal.
∴= hν0
Wavelength of red light, 
= 6328 × 10−10 m
∴Frequencyof red light, 
Since ν0> νr,the photocell will not respond to the red light produced by the laser.
Monochromatic radiation of wavelength 640.2nm (1nm = 10−9 m) from a neon lamp irradiates photosensitivematerial made of caesium on tungsten. The stopping voltage is measured to be0.54 V. The source is replaced by an iron source and its 427.2 nm lineirradiates the same photo-cell. Predict the new stopping voltage.
Answer - 25 : -
Wavelength of the monochromaticradiation, λ = 640.2 nm
= 640.2 × 10−9 m
Stopping potential of the neon lamp, V0 =0.54 V
Charge on an electron, e =1.6 × 10−19 C
Planck’s constant, h = 6.6× 10−34 Js
Let be the work function and ν bethe frequency of emitted light.We have the photo-energyrelation from the photoelectric effect as:
eV0 = hν − Wavelength of the radiation emitted from aniron source, λ‘ = 427.2 nm
= 427.2 × 10−9 m
Let 
be the new stoppingpotential. Hence, photo-energy is given as:Hence,the new stopping potential is 1.50 eV.
Question - 26 : - A mercury lamp is aconvenient source for studying frequency dependence of photoelectric emission,since it gives a number of spectral lines ranging from the UV to the red end ofthe visible spectrum. In our experiment with rubidium photo-cell, the followinglines from a mercury source were used:
λ1 = 3650 Å, λ2=4047 Å, λ3= 4358 Å, λ4= 5461Å, λ5= 6907 Å,
The stopping voltages,respectively, were measured to be:
V01 = 1.28 V, V02 =0.95 V, V03 = 0.74 V, V04 =0.16 V, V05 = 0 V
Determine the value of Planck’sconstant h, the threshold frequency and work function for thematerial.
[Note: You will notice that toget h from the data, you will need to know e (whichyou can take to be 1.6 × 10−19 C). Experiments of this kind onNa, Li, K, etc. were performed by Millikan, who, using his own value of e (fromthe oil-drop experiment) confirmed Einstein’s photoelectric equation and at thesame time gave an independent estimate of the value of h.]
Answer - 26 : -
Einstein’s photoelectricequation 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 amaterial
It can be concluded from equation (1) thatpotential V0 is directly proportional tofrequency ν.
Frequency is also given bythe relation:
This relation can be usedto obtain the frequencies of the various lines of the given wavelengths.
Thegiven quantities can be listed in tabular form as:
| Frequency × 1014 Hz | 8.219 | 7.412 | 6.884 | 5.493 | 4.343 |
| Stopping potential V0 | 1.28 | 0.95 | 0.74 | 0.16 | 0 |
The following figure shows a graphbetween νand V0.
It can be observed that the obtained curveis a straight line. It intersects the ν-axis at 5 × 1014 Hz,which is the threshold frequency (ν0) of the material. PointD corresponds to a frequency less than the threshold frequency. Hence, there isno photoelectric emission for the λ5 line, andtherefore, no stopping voltage is required to stop the current.
Slopeof the straight line = 
From equation (1), the slope 
can be written as:
The work function of themetal is given as:
= hν0
= 6.573 × 10−34 × 5 × 1014
= 3.286 × 10−19 J
=2.054 eV
The work function for thefollowing metals is given:
Na:2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will notgive photoelectric emission for a radiation of wavelength 3300 Å from a He-Cdlaser placed 1 m away from the photocell? What happens if the laser is broughtnearer and placed 50 cm away?
Answer - 27 : -
Mo and Ni will not showphotoelectric emission in both cases
Wavelength for a radiation, λ =3300 Å = 3300 × 10−10 m
Speed of light, c = 3 × 108 m/s
Planck’s constant, h = 6.6× 10−34 Js
Theenergy of incident radiation is given as:It can be observed thatthe energy of the incident radiation is greater than the work function of Naand K only. It is less for Mo and Ni. Hence, Mo and Ni will not showphotoelectric emission.
Ifthe source of light is brought near the photocells and placed 50 cm away fromthem, then the intensity of radiation will increase. This does not affect the energyof the radiation. Hence, the result will be the same as before. However, thephotoelectrons emitted from Na and K will increase in proportion to intensity.
Light of intensity 10−5 W m−2 fallson a sodium photo-cell of surface area 2 cm2. Assuming that the top5 layers of sodium absorb the incident energy, estimate time required forphotoelectric emission in the wave-picture of radiation. The work function forthe metal is given to be about 2 eV. What is the implication of your answer?
Answer - 28 : -
Intensity of incident light, I =10−5 W m−2
Surface area of a sodium photocell, A =2 cm2 = 2 × 10−4 m2
Incident power of the light, P = I ×A
= 10−5 × 2 × 10−4
= 2 × 10−9 W
Workfunction of the metal, = 2 eV= 2 × 1.6 × 10−19
= 3.2 × 10−19 J
Number of layers of sodium that absorbs theincident energy, n = 5
We know that the effective atomic area of asodium atom, Ae is 10−20 m2.
Hence, the number of conduction electronsin n layers is given as:
The incident power isuniformly absorbed by all the electrons continuously. Hence, the amount ofenergy absorbed per second per electron is:
Time required forphotoelectric emission:
Thetime required for the photoelectric emission is nearly half a year, which isnot practical. Hence, the wave picture is in disagreement with the givenexperiment.
Question - 29 : - Crystal diffraction experiments can beperformed using X-rays, or electrons accelerated through appropriate voltage.Which probe has greater energy? (For quantitative comparison, take thewavelength of the probe equal to 1 Å, which is of the order of inter-atomicspacing in the lattice) (me= 9.11 × 10−31 kg).
Answer - 29 : -
An X-ray probe has agreater energy than an electron probe for the same wavelength.
Wavelength of light emitted from theprobe, λ = 1 Å = 10−10 m
Mass of an electron, me =9.11 × 10−31 kg
Planck’s constant, h = 6.6× 10−34 Js
Charge on an electron, e =1.6 × 10−19 C
The kinetic energy of theelectron is given as:
Where,
v = Velocity of the electron
mev = Momentum (p) of the electron
According to the deBroglie 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.
Question - 30 : - (a) Obtain the deBroglie wavelength of a neutron of kinetic energy 150 eV. As you have seen inExercise 11.31, an electron beam of this energy is suitable for crystaldiffraction experiments. Would a neutron beam of the same energy be equallysuitable? Explain. (mn= 1.675 × 10−27 kg)
(b) Obtain the deBroglie wavelength associated with thermal neutrons at room temperature (27ºC). Hence explain why a fast neutron beam needs to be thermalised with theenvironment before it can be used for neutron diffraction experiments.
Answer - 30 : - (a) DeBroglie wavelength =
; neutron is not suitable for thediffraction experiment
Kinetic energy of the neutron, K =150 eV
= 150 × 1.6 × 10−19
= 2.4 × 10−17 J
Mass of a neutron, mn =1.675 × 10−27 kg
The kinetic energy of theneutron is given by the relation:
Where,
v = Velocity of the neutron
mnv = Momentum of the neutron
De-Broglie wavelength ofthe neutron is given as:
It is given in the previous problem that theinter-atomic spacing of a crystal is about 1 Å, i.e., 10−10 m.Hence, the inter-atomic spacing is about a hundred times greater. Hence, aneutron 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 energyof the neutron is given as:
Where,
k = Boltzmann constant = 1.38 × 10−23 Jmol−1 K−1
The wavelength of theneutron is given as:
Thiswavelength is comparable to the inter-atomic spacing of a crystal. Hence, thehigh-energy neutron beam should first be thermalised, before using it fordiffraction.