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 Bihar Board Class 12th Physics Solutions Chapter 11 Dual Nature of Radiation and Matter

Bihar Board Class 12 Physics Dual Nature of Radiation and Matter Textbook Questions and Answers

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kVelectrons.
Answer:
Here, V = potential difference = 30 kV = 30,000 V.
(a) Let v max = maximum frequency of X-rays produced = ?
Using the formula E = hv = eV, we get
v max = eVh
Here, h = 6.62 x 10-34Js,
e = 1.6 x 10-19 C.
v max = 1.6×10−19×3×1046.62×10−34
= 7.24 x 10-18 Hz.

(b) Let λmin = minimum wavelength of X-rays produced = ?
Using the relation, c = vλ., we get
λmin = cvmax
where c = 3 x 108 ms-1, vmax = 7.24 x 10-18 Hz
λmin = 3×1087.24×1018 = 4.14 x 1011 m
= 4.14×10−1010 = 0.414 x 10-10m
= 0.414 Å = 0.0414 nm (∴1 Å = 10-10 m)

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 x 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons.
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Answer:
Here, ω = work function of Caesium metal = 2.14 eV = 2.14 x 1.6 x 10-19J.
v = frequency of incident light = 6x 10-14Hz.
h = Planck’s constant = 6.62 x 10-34 Js.
∴ Energy of incident light = hv = 6.62 x 10-34 x 6 x 1014 J.

(a) Emax = maximum kinetic energy of photo electrons emitted = ?
Using Einstein’s photoelectric equation, hv = ω + Emax, we get
Emax = hv – w

(b) V0 = Stopping potential =?
By definition, eV0 = Emax

(c) Let vmax = maximum speed of emitted photo electron = ?
M = mass of electron = 9.1 x 10-31 kg
Using formula,

Question 3.
The phtoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Here, VQ = cut off voltage or stopping potential = 1.5 V
Emax= Maximum Kinetic energy emitted = ?
e = charge of an electron = 1.6 x 10-19 C
Using the relation,
KE = Emax = eV0we get
Emax = 1.6 X 10-19 x 1.5 = 2.4 x 10-19J

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assuume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the proton?
Answer:
Here, X = wavelength of monochromatic light = 632.8 nm = 632.8 x 10-9 m ,
P = Power emitted by helium-neon laser = 9.42 mW = 9.42 x 10-3 W.

(a) E = energy of each photon = ?
p = momentum of each photon = ?, h = 6.62 x 10-34 JS,
c = 3 x 108 ms-1.
Using relation, E = hv = hcλ, we get

(b) Let n = Number of photons emitted per second = No. of photons reaching the target/sec = ?
∴ Energy emitted by these photons/sec. = nhv = nE.
Also P = energy emitted/sec = 9.42 x 10-3 W.
∴ nE = P
or n = pE = 9.42×10−33.14×10−19 = 3 x 1016

(c) Let v = speed of hydrogen atom = ?
m = mass of hydrogen atom = mass of proton = 1.67 x 10-27kg.
Let p’ = momentum of hydrogen atom = mv
∴ As per statement
p’ = p = momentum of photon
= 1.05 x 10-27 kg ms-1,
or mv = 1.05 x 10-27
∴ v = 1.05×10−271.67×10−27 = 0.628 ms-1
= 0.63 ms-1

Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 x 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:
Here, λ = wavelength of photons in the sunlight = 550 nm = 550 x 10-9 m.
ø = Energy flux = Energy of number of photons from sun light reaching earth per second per unit area
= 1.388 x 103 W/m2 = 1.388 x 103 J s-1 m-2 …….(1)
[∴ P = Et
∴ W = 1 Js-1]
h = plank’s constant – 6.62 x 1034 Js.
c = 3 x 108ms-1.
Let n = number of photons incident on earth per sec per m2 = ?
Let E’ = energy of each photon = hv = hcλ
If E be the energy of all photons reaching the surface of earth per sec. per unit area, then
E = nE’ = nhv = nhcλ ………….(2)
∴ From (1) and (2), we get nhc

Question 6.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10-15 V s. Calculate the value of Planck’s constant.
Answer:
Here, slope of cut-off voltage versus frequency curve = tan θ = Vv = 4.12 x 10-15Vs
Where θ = angle made by the curve with the frequency axis, h = plank’s constant = ?, e = 1.6 x 10-19 C.
Also we know that

Question 7.
A100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Answer:
Here, P = power of sodium lamp = 100 W λ = wavelength of each photon = Wavelength of sodium light = 589 nm = 589 x 109 m.
∴Energy radiated per second by the lamp = 100 JS-1. (∴P = Et)
h = Plank’s constant = 6.62 x 10-34 JS, c = 3 x 108 ms-1.

(a) Energy per photon = E = ?
Using formula, E = hv = hcλ, we get

(b) Let n = rate at which the photons are delievered to the sphere = no. of photons delivered to sphere per sec = ?

Question 8.
The threshold frequency for a certain metal is 3.3 x 1014 Hz. If light of frequency 8.2 x 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:
Here, v0 = threshold frequency = 3.3 x 1014 Hz.
v = frequency of incident light photons = 8.2 x 1014 Hz.
V0 = cut-off voltage for photoelectric emission = ?
h = 6.62 x 10-34 JS; e = 1.6 x 10-19 C.
Using the relation, hv – hv0 = eV0, we get

Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer:
Here, W = Work function of certain metal = 4.2 eV = 4.2 x 1.6 x 10-19 J = 6.72 x 10-19 J.
λ = Wavelength of incident radiation = 330 nm = 330 x 10-9 m.
If E be the energy of the photon of incident light, then

Now from equation (1) and (2), it is clear that the frequency of
incident radiation y is less than the threshold frequency y0, hence
photoelectric emission for this radiation cannot take place.

Question 10.
Light of frequency 7.21 x 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Here, v’= frequency of incident light = 7.21 x 1014 Hz.
υmax = maximum speed of electrons ejected from the surface = 6.0 x 105 ms-1.
m = mass of electrons = 9.1 x 10-31 kg
v0 = threshold frequency = ?
h = Plank’s constant = 6.62 x 10-34 JS.
Using Einstein’s photo electric equation, hv – hv0 = 12 mu2max , we get

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made.
Answer:
Here, λ = wavelength of light = 488 nm = 488 x 10-9 m.
V0 = Stopping potential = 0.38 V
W = Work function of the material = ?
c = 3 x 108 ms-1.
h = Plank’s constant = 6.62 x 10-34 JS.
e = 1.6 x 10-19 C.
∴Energy of incident photon = λv = hvλ = 6.62×10−34×3×108488×10−9 = 4.08 x 10-19 J.
Using formula, eV0 = hv – W, we get
W = hv – eV0 = 4.08 x 10-19J – 1.6 x 10-19 x 0.38
= (4.08 – 0.608) x 10-19 J
= 3.472 x 10-19 J
= 3.472×10−191.6×10−19 eV = 2.17 eV

Question 12.
Calculate the
(a) momentum, and
(b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Here, V = P.D. applied = 56 V

(a) p = momentum of electron = ?
We know that 12 mυ2 = eV
∴ υ = 2eVm−−−√, where m, υ are the mass and velocity of the electron.
m = 9.1 x 10-31 kg, e = 1.6 x 10-19 C.
Using p = mv, we get
p = m = 2meV−−−−−√ we get
p = 2×9.1×10−31×1.6×10−19×56−−−−−−−−−−−−−−−−−−−−−−−−−−−√ = 1.631×10−47−−−−−−−−−−−√
or p = 4.04 x 10-24 kg ms-1,

(b) λ = de Broglie wavelength of the electron = ?
λ = 12.27V√ Å we get
λ = 12.2756√ Å = 12.277.483Å = 1.64 Å
= 1.64 x 10-10 m
= 0.164 x 10-9m
= 1.64 x 10 nm

Question 13.
What is the
(a) momentum.
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer:
Here, E = Kinetic energy of electron = 120 eV
= 120 x 1.6 x 10-19 J = 1.92 x 10-17J
m = mass of electron – 9.1 x 10-31 Kg

(a) h = 6.62 x 10-34 JS.
momentum of electron = p = ?
Using relation, E = p22 m, we get [∵E =12mv2=m2v22 m=p22 m]
p = 22mE−−−−−√
= 2×9.1×10−31×1.92×10−17−−−−−−−−−−−−−−−−−−−−−−−−√
= 5.91 x 10-24 kg ms-1,

(b) Let v = speed of electron = ?
Using relation, p = mv, we get
v = pm = 5.91×10−249.1×10−31 = 6.5 x 106 ms-1

(c) de-Broglie wavelength is given by
λ = hp = 6.62×10−345.91×10−24 = 1.12 x 10-10 m
= 0.112 x 10-9m
= 0.112 nm

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de- Broglie wavelength.
Answer:
Here, X = wavelength of light – 589 nm – 589 x 10-9 m.
me= mass of electron = 9.1 x 10-31 kg.
mn = mass of neutron = 1.67 x 10-27 kg.
h = Plank’s constant = 6.62 x 10-34 JS.

(a) Let E1 = K.E. of electron = ?
Using the relation λ = h2meE√, we get
λ2 = h22 meE

Question 15.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s.
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 x 10-9 kg drifting with a speed of 2.2 m/s?
Answer:
(a) Here, mb = mass of bullet = 0.040 kg
vb = velocity of bullet = 1.0 km s-1 = 1000 ms-1,
h = 6.62 x 10-34 JS.
λb = de Broglie wavelength of bullet = ?
Using formula, λ = hmv, we get
λb = hmbvb=6.62×10−340.040×1000
= 1.655 x 10-35 m = 1.7 x 10-35 m

(b) Here, m = 0.060 kg
v = 1.00 ms-1
λ = ?
∴ λ = hmv=6.62×10−3460×10−3×1.0 = 1.1 x 10-32 m

(c) Here,
m = 1.0 x 10-9 kg
v = drift speed of dust particle
= 2.2ms-1
λ = deBroglie wavelength = ?
p = mv =10 x 10-9 x 2.2
= 2.2 x 10-9 kg ms-1
λ = hmv = hp = 6.62×10−342.2×10−9 = 3.01 x 10-25 m.

Question 16.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer :
Here λc = wavelength of an electron = 1.0 nm = 10-9 m.
λp = wavelength of a photon = 10-9 m.
h = 6.63 x 10-34Js, c = 3 x 108 ms-1

(a) (i)Momentum of electron = pe
Using formula, λ = hp, we get

(b) Let Ep = Kinetic energy of photon = ?
pe = hλe = 6.63×10−3410−9 = 6.63 x 10-25 kg ms-1

(ii) Let pp = momentum of photon
∴ pp = hλp=6.63×10−3410−9 = 6.63 x 10-25 kg ms-1

(b) Let Ep = kinetic enegry of photon = ?
Using formula, E = hv . hcλ , we get

(c) Ee = K.E of electron
Using formula,

Question 17.
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 x 10-10 m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of 32 k T at 300 K.
Answer:
(a) Here, λ = deBroglie wavelength = 1.40 x 10-10 m.
h = 6.62 x 10-34 Js,
E = Kinetic energy of neutron = ?
m = mass of neutron = 1.67 x 10-27kg.

Using formula, λ = h2mE√, we get

(b) Here, T = absolute Temperature = 300 K
k = Boltzman’s constant
= 1.38 x 10-23 JK-1.
E = Kinetic energy of neutron = 32 kT
= 32 x 1.38 x 10-23 x 300
= 6.21 x 10-21 J.
m = mass of neutron = 1.67 x 10-27 kg.
λ = deBroglie wavelength of neutron = ?

Using the relation,

Question 18.
Show that the wave length of electromagnetic radiation is equal to the de Brogue wave length of its quantum.
Answer:
Let v, λ’ be the frequency and wave length of the photon having momentum
∴ p = EC=hvC
Where C = speed of photon
= Speed of electromagnetic radiation
∴ de Brogue wavelength A’ of the photon i.e. quantum of e.m. wave is given by
λ’ = hp=h(hvC)
= hh2=Cv ……….(1)
Also let A be the wavelength of electromagnetic radiation of frequency V.
∴ λ=Cv ……….(2)
where C = speed of photon
= speed of electromagnetic radiation.
∴ From (1) and (2)
λ = λ’ = Cλ
Hence Proved.

Question 19.
What is the de Brogue wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atonlic mass of nitrogen = 14.0076 u)
Answer:
Here, k BoltzmaWs constant = 1.38 x 10-23 JK-1.
T = absolute temp = 300 K
m = moLecular weight of Nitrogen molecule = 28.0152 u
= 28.0152 x 1.67 x 10-27 Kg
Let C = r.m.s. speed of N2 molecule at 300 K.
∴ According to Kinetic Theory of gases, K.E./molecule = 32 kT or C = 3kTm−−−√
C = 3×1.38×10−23×30028.0152×1.67×10−27−−−−−−−−−−−−√
= 5.15 x 102 ms-1
λ = deBroglie wavelength = ?

Using formula, λ = hmC
= 6.62×10−3428.0152×1.67×10−27×5.15×102
= 2.75 x 1011 m
= 0.0275 x 109 m
= 0.0275 nm.
= 0.028 nm

Question 20.
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The ‘specific charge’ of the electron i.e., its – is given to be 1.76 x 1011 C kg-1.

(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Answer:
(a) Here, V = potential difference between cathode and anode = 500 V.
em = specific charge of electron = 1.76 x 1011 C kg-1.
υ = speed of emission of electrons from cathode = ?
Using the relation, eV = 12 mv2 …………..(1)
we get
υ = 2emV−−−−√ ……….(2)
or υ = 2×1.76×1011×500−−−−−−−−−−−−−−−−−√
= 1.33 x 107 ms-1,

(b) Here, V = anode potential = 10 MV = 10 x 106 V = 107 V.
Using equation (2), υ = 2emV−−−−√
υ = 2×1.76×1011×107−−−−−−−−−−−−−−−−−√
= 1.88 x 109 ms-1
This velocity υ is greater than the velocity of light C in free space (which is 3 x 108 ms-1). No material particle can have velocity comparable to C. Therefore, K.E. = 12 mv2 is not exact relation.
So, the formula of K.E. is to be modified by taking into consideration special theory of relativity. Thus the relativistic expression
for K.E. replacing 12 mv2 becomes mc2 – mQ c2 = c2 (m – m0) where m0 is the rest mass of an electron = 9 x 10-31 kg and m is relativistic mass of electron given by
m = m01−v2C2√
∴ eqn. (1) is to be modified as :
(m – m0) c2 = eV

Squaring on both sides, we get

i. e. v is now comparable to c.

Question 21.
(a) A monoenergetic electron beam with electron speed of 5.20 x 106 ms-1 is subject to a magnetic field of 1.30 x 10-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given – for electron equals 1.76 x 1011 C kg-1.

(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
[Note : Exercises 11.20 (b) and 11.21 (b) take you to relativistic mechanics which is beyond the scope of this book. They have beer, inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answer at the end to know what ‘very high speed or energy means.]
Answer:
Here, υ = speed of electrons beam = 5.20 x 106 ms-1.
B = magnetic field = 1.30 x 10-4 T
em = specific charge of electron = 1.76 x 1011 C kg-1. m
θ = angle between B⃗  and the υ⃗  = 90°
Force exerted by B⃗  on the electron is
F¯m = e(v⃗ ×B⃗ )
Fm = e υ B sin 90° = eυB
This force provides the necessary centrepetal force for tracing out a circle i.e.,

(b) Here, E = energy = 20 MeV= 20 x 1.6 x 10-13J
Using relation, E = 12 mv2, we get
υ = 2Em−−−√=2×20×1.6×10−139.1×10−31−−−−−−−−−−√
= 2.67 x 109 ms-1
which is greater than the velocity of light.
So the formula used in case (a) i.e., r = mveB is not valid for calculating the radius of the path of 20 MeV electron beam because electron with such a high energy has velocity in the relativistic domain i.e., comparable with the velocity of light and the mass varies with the increase in velocity but we have taken it as constant.
∴m = m01−v2c2√ is to be considered.
Thus the modified equation becomes :
r = ⎛⎝⎜m01−v2c2√⎞⎠⎟veB

Question 22.
An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10-2mm of Hg). A magnetic field of 2.83 x 10-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture : this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.
Answer:
Here, V = potential at anode = 100 V
B = magnetic field = 2.83 x 10-4 T
radius of circular path,
r = 12 cm = 0.12 m
em = ?
Let υ = Velocity acquired by electrons
m = mass of electron
∴Gain in K.E. of electrons when accelerated through V volts is given by
12mv2 = eV
or mv2 = 2eV
As the electrons move in the circular path so
magnetic force = Centrepetal force
or eυB = mv2r
or v = eBrm ………..(ii)
∴ From (i) and (ii), we get
m (eBrm)2 = 2eV
or em = 2 V B2r2
= 2×100(2.83×10−4)2×(0.12)2
= 1.73 x 1011 C Kg1

Question 23.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 A. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Answer:
Here, λ min = 0.45 A = 0.45 x 1010 m.
Emax = maximum energy of X-ray photon = ?
h = Plank’s constant = 6.62 x 10-34 JS.
c = Velocity of light = 3 x 108 ms-1.

Using Formula,

(b) In X-ray tube, accelerating voltage provides the energy to the electrons which produce X-rays. For getting X-ray photons of 27.61 KeV, it is required that the incident electrons must possess kinetic energy of atleast 27.61 KeV. The order of the accelerating voltage can be calculated as :
Emax = eV
V = Emaxe=4.42×10−51.6×10−19 V
= 27.61 x 103 V = 27.61 KV.
Order of the accelerating voltage is 30 KV.

Question 24.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1 BeV = 109 eV)
Answer:
Here, Energy of 2γ-rays = 10.2
BeV = 10.2 x 109 eV
(∴1 BeV = 109 eV).
∴ Energy of each y-ray is given by
E = 12 (10.2 x 109 eV) = 12 x 10.2 x 109 x 1.6 x 10-19 J
= 8.16 x 10-10 J.
λ = wavelength of each γ-ray = ?
Using formula, E = hcλ, we get
λ = hcE = 6.62×10−34×3×1088.16×10−10
= 2.44 x 10-16 m.

Question 25.
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~10-10 Wm-2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 x 1014 Hz.
Answer:
(a) Here, P = power of transmitter = 10 kVV = 104 W
λ = wavelength of radiowaves = 500 m.
h = 6.626 x 10-34 JS,
c = 3 x 108 ms-1.
Let E = energy of each photon
∴ E = hv = hcλ = 6.626×10−34×3×108500 = 3.98 x 10-28 J
n = no. of photons emitted per second = ?
n is given by

We see that the energy of a radio photon is very small and the number of photons emitted per second in a radio beam is very large. So there is negligible error involved in ignoring the energy of photon and we can treat total energy of radio waves as continuous.

(b) Here, A = area of pupil = 0.4 cm2 = 0.4 x 10-4 m2.
v = average frequency of white light = 6 x 1014Hz
I = minimum intensity of white light = 10-10Wm-2.
∴ E = energy of each photon = hv = 6.626 x 10-34 x 6 x 10-14 = 3.98 x 10-19 J.
Let n = no. of photons falling per sec. per unit area.
∴ energy per unit area per sec. due to these photons.
= total enrgy of n photons.
= nE = n x 3.98 x 10-19 Js-1m-2
Also I = energy per sec. per unit area.
∴ 10-10 = n x 3.98 x 10-19
or n = 10−113.98×10−14 = 2.51 x 108 m-2s-1
∴ No. of photons entering the pupil per second
n x area of pupil
= 2.51 x 108 x 0.4 x 10-4 ≈ 104 s-4.
Though this number is not as large as in (a). but still large enough to be cou nted
Comparison of case (a) and (b) tells that our eye cannot count the number of photons individually.

Question 26.
Ultraviolet light of wavelength 2271 A from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is – 1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~ 10-5 W m-2) red light of wavelength 6328 Å produced by a He-Ne laser?
Answer:
Here, Power P = 100 W
Stopping potential, V0 = 1.3 V (magnitude)
λr = wavelength of red light = 6328 Å = 6328 x 10-10 m. h “ plank’s constant – 6.62 x 108ms-1 Js, c = 3 x 108 ms-1.
h = wavelength of UV light = 2271 A = 2271 x 10-10 m.
λu = work function of metal = ?
Using Einstein’s photoelectric equation,

Let v0 = frequency for red light.
∴ hv0 = W
or v0 = Wh=4.17×1.6×10−196.62×10−34
= 1.0 x 1015 Hz
Let vr = frequency for red light.
vr = cλr = 3×1086328×10−10 = 4.74 x 1014Hz
Since vr < v0 i.e., frequency produced by He-Ne laser incident on the photocell is lesser than the threshold frequency, so the photocell will not respond.

Question 27.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 109 m) from a neon lamp irradiates photosensitive material • made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:
Here, for neon lamp; λ = wavelength = 640.2 nm
= 640.2 x 10-9 m.
V0 = Stopping potential = 0.54 V.
h = 6.62 x 1034 Js,
c = 3 x 108 ms-1 .
W = Work function = ?
Using formula,

For iron source :
λ = wavelength = 427.2 nm
= 427.2 x 10-9 m
W = work function = 2.238 x 10-19J
V0 = stopping potential = ?
From eqn. (1), we get

Question 28.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :
λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4= 5461 Å, 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 = 0V
(a) Determine the value of Planck’s constant h the threshold frequency and work function for the material.
[Note : You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 x 10 19 C). Experiments of this kind on Na, Li, K etc. were performed by Milikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h ]
Answer:
From Einstein photoelectric equation,
eV0 = hv – W or hv – eV0 = W
or hcλ – eV0 = W ………(1)
For λ = λ1, V0 = V01, we get
hcλ1 – eV01 = W ………..(2)
and for λ = λ2, V0 = V02, we get
hcλ2 – eV02 = W …………(3)
(2) – (3) gives

From (5), (7) and (9), we conclude that the approximate value of h is 6.6 x 10-34 Js.
From eqn. (2), we get

Question 29.
The work function for the following metals is given : Na : 2.75 eV; K : 2.30 eV; Mo : 4.17 eV; Ni : 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 A from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Answer:
Here, for all metals wave length of incident radiation,
λ = 3300 Å = 3300 x 10-10 m.
∴ Frequency of incident radiation is given by :
v = cλ = 3×1083300×10−10 = 9.1 x 1014Hz
If v0 be the threshold frequency, then
hv0 = W
v0 = Wh ……….(1)
Also it is given that

Now we see that the frequency of incident radiation is greater than the threshold frequency of Na and K metals, but lesser than those of Mo and Ni. So thephotoelectric emission will take place only in Na and K metals and not in Mo and Ni.
Alliter :
Energy of incident radiation is –

which is greater than W of Na and K and lesser than that for Mo and Ni. So emission in Na and K will take place. Now r = 1 m, r’ = 50 cm = 0.50 m.
If the laser is brought closer, the intensity of incident radiation increases. This does not affect the result regarding Mo and Ni metals, while photoelectric emission from Na and K will increase in proportion to intensity as I ∝ 1r2 and becomes 4 times when r’ = r2

Question 30.
Light of intensity 10-5 W m-2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Answer:
Here; I = intensity of incident light = 105 Wm-2.
A = surface area of each layer in sodium = 2 cm2 = 2 x 10-4 m2.
W = Work function for metal = 2eV = 2 x 1.6 x 10-19 J.
Since r = size of atom is about 10-10 m (= 1 Å), so the effective area of an atom of Na may be considered as ≈ r2 = 10-20 m2 (roughly)
∴ Number of atoms in the 5 layers of sodium =

We know that sodium has one free electron (or conduction electron) per atom.
∴ No. of electrons in 5 layers of sodium = 1017. Now the incident power on the surface area of the photocell, P = IA = 102 x 2 x 10-4 = 2 x 10-9 W.

Since incident energy is absorbed by 5 layers of sodium, so according to wave picture of radiation, the electron present in all the 5 layers of sodium will share the incident energy equally.
∴ energy recieved by any one electron in 5 layers of sodium =

Since the work function of sodium is 2 eV, an electron w.ill be ejected as soon as it gathers energy’ equal to 2eV.
∴ time required for photoelectric emission
WE=3.2×10−192×10−26
= 1.6 x 107s ≈ 0.5years
Implication : The answer obtained implies that the time of emission of electron is very large and is not in agreement with the observed time of emission, which is nearly 10″9 s i.c., there is no time lag between the incidence of light and the emission of photo electrons. Thus we conclude that wave-picture of radiation is not applicable for photo-electric emission.

Question 31.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to lA, which is of the order of interatomic spacing in the lattice) (me = 9.11 x 10-31 kg).
Answer:
Here, λ = wavelength of the probe = 1 Å = 10-10m.
me = mass of electron = 9.11 x 10-31 kg.
h = Plank’s constant = 6.62 x 10-34 JS.
c = 3 x 108 ms-1
For X-ray photon of 1Å – If Ex be the energy of X-ray photon, then
Using formula, E = hcλ, we get

For electron of wavelength 1Å
Using formula, λ = hmυ, we get
mυ = hλ=6.62×10−3410−10
= 6.62 x 10-24 kg ms-1
if E be the enegry of electron, then

Clearly X-ray photon has much greater energy than energy of electron for same wavelength.

Question 32.
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn = 1.675 x 10-27 kg).
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer:
(a) Here, mn = mass of neutron = 1.675 x 10-27 kg
En = Kinetic energy of neutron = 150 eV = 150 x 1.6 x 10-19 J = 2.4 x 10-17 J.
λn = wavelength of neutron = ?
Using Formula,

This wavelength is about hundred times smaller than the interatomic spacing (≈ 10-10 m) of the crystals, thus a neutron of 150 eV energy is not suitable for crystal diffraction experiments.
(b) Here, T = 27°C = 273 + 27 = 300 K
mn = 1.675 x 10-27 kg,
k = 1.38 x 10-23J k-1 molecule-1
E = energy of neutron at temperature T = 32 kT
λ = wavelength of thermal neutrons = ?
Using formula,

This wavelength is comparable to the interatomic spacing of crystals (≈ 10-10 m). So thermal neutrons are able to interact with the crystal and hence are suitable for diffraction experiments. Since λ ∝ 1T√, so increasing their temperature, decreases their de Broglie wavelength and they become unsuitable for crystal diffraction. Fast neutrons possess wavelength quite small as compared to interatomic spacing. Hence they need to be thermalised with the environment for neutron diffraction experiments.

Question 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:
Here, V = acceleration potential = 50 KV = 50 x 103 V.
m = 9.1 x 10-31 kg,
e = 1.6 x 10-19 C,
h = 6.62 x 10-34 Js.
λe = wavelength of electron = ?
E = energy of electrons is given by
E = 12 mv2 = eV
or E = eV = 1.6 x 10-19 x 50 x 103J
= 80 x 10-16 J

Using formula,
λ = h2mE√, we get , λe = h2mE√

Also for yellow light λy = 5990 Å x 10-10m.
Also we know that the resolving power of a microscope is inversely proportional to the wavelength of the radiation used.
i.e R.P ∝1λ

i.e R.P. of elecron Microscope is 105 times larger than the R.P of optical Microscope

Question 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:
Here, λ = size of structure = 10-15 m.
h = 6.62 x 10-34 Js,
c = 3 x 108 ms-1.
m0 C2 = rest mass energy = 0.511 MeV
= 0.511 x 1.6 x 10-13 J = 8.18 x 10-14 J.
∴ Using the relation, λ = hp, we get
P = hλ = 6.62×10−3410−15
= 6.62 x 10-19 kg ms-1.
Using the relativistic formula for total energy of particle, we get

Thus energy of the proton ejected out of the linear accelerator is of the order of BeV.

Question 35.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and l atm pressure; and compare it with the mean separation between two atoms under these conditions.
Here, t = 27°C
Atomic mass of He = 4g
T = 273 + 27 = 300 k
N = Avogadro’s number
h = 6.62 x 10-34sec. = 6 x 1023
Let m = mass of He atom

Let r0 be the mean separation between the two atom = ?
P = 1 atom = 1.01 x 105 Pascal/m2
Using the relation, PV = RT + b NT (∴ R = kN)
∴ VN=k TP
∴Mean separation between two atom is given by :
r0 = (vN)1/3

From (1) and (2), we see that so is much logner than de Broglie wave length.
i.e λr0=0.73×10−1034×10−10 = 0.021 m

Question 36.
Compute the typical de Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 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:
Here t = 27° C
∴ T = 27 + 273 = 300 K.
M = mass of an electron = 9.1 x 10-31 kg
λ = de Broglie wavelength = ?
λ = 6.62 x 10-34sec
k = Boltzmann’s contant = 1.38 x 10-23JK-1 mol-1
r0 = mean separation between two electrons in a metal = 2 x 10-10 m = 2 A.
K.E. of electOm at T is given by
E = 32 KT.
de Broglie wave length of electron is given by

i.e. λ is 31 times the inter-electron separation (r0) in a metal. Thus from Q. 35 and Q. 36, we conclude that electron wave packets. In a metal strongly overlap with one another while the wave packets associated with gaseous molecules under ordinary conditions don’t overlap.

Question 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 ?
Answer:
(a) The quarks having fractional charges are thought to be confined within a proton and a neutron. These quarks are bound by forces. These forces become stronger when the quarks are tried to be pulled a part. That is why, the quarks always remain together. It is due to this reason that though fractional charges does not exist in nature but the observable charges are always integral multiple of charge of electrons.

(b) What is so special about the combination em not simply talk of e and m separately?
Answer:
Both the basic relations eV = 12 mV2 or eE = ma and eBv = mV2r for electric and magnetic fields respectively show that the dynamics of electrons is determined not by and m separately but by the combination of em

(c) Why should gases be insulatprs at ordinary pressures and start conducting at very low pressures?
Answer:
At low pressure, ions have a chance to reach their respective electrodes and constitute a current. At ordinary pressures, ions have no chance to db so because of collisions with gas molecules and recombination.

(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?
Answer:
Work function merely indicates the minimum energy required for the electron in the highest level of conduction band to get out of the metal. Not all electrons in the metal belong to this level. Consequently, for the same incident radiation, electron knocked off from different levels come out with different energies.

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the radiations :
E = hv, p = hλ
But while the value of A is physically significant, the value of v(and therefore, the value of the phase speed vλ) has no physical significance. Why?
Answer:
The absolute value of energy E (but not momentum) of any particle is arbitrary to within an additive constant. Hence, while λ is physically significant, absolute value of v of matter wave of an electron has no direct physical meaning. The phase speed vλ is likewise not physically significant. The group speed given by
dvd(1λ)=dEdp
= ddp(p22m)=pm
is physically meaningful.