Chapter 11 Dual Nature Of Radiation And Matter Solutions
Question - 1 : - Find the
(a) maximum frequency,and
(b) minimum wavelength of X-rays produced by 30 kVelectrons.
Answer - 1 : -
Potential of 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 × 10−19 C
(a)Maximum frequency producedby the X-rays = ν
The energy of theelectrons is given by the relation:
E = hν
Where,
h =Planck’s constant = 6.626 × 10−34 Js
Hence, the maximum frequency of X-rays producedis
(b)The minimum wavelengthproduced by the X-rays is given as:
Hence,the minimum wavelength of X-rays produced is 0.0414 nm.
The work function of caesium metal is 2.14eV. When light of frequency 6 ×1014 Hz is incident on the metalsurface, photoemission of electrons occurs. What is the
(a) maximum kineticenergy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of theemitted photoelectrons?
Answer - 2 : - Work function of caesium metal, 
Frequency of light, 
(a)The maximum kinetic energyis given by the photoelectric effect as:
Where,
h = Planck’s constant = 6.626 × 10−34 Js
Hence, the maximum kineticenergy of the emitted electrons is 0.345 eV.
(b)For stopping potential
, we can write the equation forkinetic energy as:
Hence, the stoppingpotential of the material is 0.345 V.
(c)Maximum speed of theemitted photoelectrons = v
Hence,the relation for kinetic energy can be written as:
Where,
m = Mass of an electron = 9.1 × 10−31 kg
Hence,the maximum speed of the emitted photoelectrons is 332.3 km/s.
Thephotoelectric cut-off voltage in a certain experiment is 1.5 V. What is themaximum kinetic energy of photoelectrons emitted?
Answer - 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 × 10−19 C
Therefore,the maximum kinetic energy of the photoelectrons emitted in the givenexperiment is 2.4 × 10−19 J.
Monochromatic light ofwavelength 632.8 nm is produced by a helium-neon laser. The power emitted is9.42 mW.
(a) Find the energy andmomentum of each photon in the light beam,
(b) How many photons persecond, on the average, arrive at a target irradiated by this beam? (Assume thebeam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel inorder to have the same momentum as that of the photon?
Answer - 4 : -
Wavelength of the monochromatic light, λ =632.8 nm = 632.8 × 10−9 m
Power emitted by the laser, P =9.42 mW = 9.42 × 10−3 W
Planck’s constant, h =6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
Mass of a hydrogen atom, m =1.66 × 10−27 kg
(a)The energy of each photonis given as:
The momentum of eachphoton is given as:
(b)Number of photons arrivingper second, at a target irradiated by the beam = n
Assume that the beam has auniform cross-section that is less than the target area.
Hence, the equation forpower can be written as:
(c)Momentumof the hydrogen atom is the same as the momentum of the photon,
Momentum is given as:
Where,
v = Speed of the hydrogen atom
Question - 5 : - The energy flux of sunlight reaching thesurface of the earth is 1.388 × 103 W/m2. How manyphotons (nearly) per square metre are incident on the Earth per second? Assumethat the photons in the sunlight have an average wavelength of 550 nm.
Answer - 5 : -
Energy flux of sunlight reaching the surfaceof earth, Φ = 1.388 × 103 W/m2
Hence, power of sunlight per squaremetre, P = 1.388 × 103 W
Speed of light, c = 3 × 108 m/s
Planck’s constant, h =6.626 × 10−34 Js
Averagewavelength of photons present in sunlight, 
Number of photons per square metre incidenton earth per second = n
Hence, the equation forpower can be written as:
Therefore, every second, 
photons are incident per square metre on earth.
In an experiment on photoelectric effect,the slope of the cut-off voltage versus frequency of incident light is found tobe 4.12 × 10−15 V s. Calculate the value of Planck’s constant.
Answer - 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 × 10−19 C
h = Planck’s constant
Therefore, the value of Planck’s constantis 
Question - 7 : - A100 W sodium lamp radiates energy uniformly in all directions. The lamp islocated at the centre of a large sphere that absorbs all the sodium light whichis incident on it. The wavelength of the sodium light is 589 nm. (a) What isthe energy per photon associated with the sodium light? (b) At what rate arethe photons delivered to the sphere?
Answer - 7 : -
Power of the sodium lamp, P =100 W
Wavelength of the emitted sodiumlight, λ = 589 nm = 589 × 10−9 m
Planck’s constant, h =6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
(a)The energy per photonassociated with the sodium light is given as:
(b)Number of photonsdelivered to the sphere = n
Theequation for power can be written as:
Therefore, every second,
photons are delivered to the sphere.
Question - 8 : - The threshold frequency for a certain metalis 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hzis incident on the metal, predict the cutoff voltage for the photoelectricemission.
Answer - 8 : - Threshold frequency of the metal, 
Frequency of light incident on the metal, 
Charge on an electron, e =1.6 × 10−19 C
Planck’s constant, h =6.626 × 10−34 Js
Cut-offvoltage for the photoelectric emission from the metal = The equation for thecut-off energy is given as:
Therefore, the cut-off voltage for thephotoelectric emission is 
Thework function for a certain metal is 4.2 eV. Will this metal give photoelectricemission for incident radiation of wavelength 330 nm?
Answer - 9 : -
No
Workfunction of the metal, 
Charge on an electron, e =1.6 × 10−19 C
Planck’s constant, h =6.626 × 10−34 Js
Wavelength of the incident radiation, λ =330 nm = 330 × 10−9 m
Speed of light, c = 3 × 108 m/s
The energy of the incidentphoton is given as:
.Itcan be observed that the energy of the incident radiation is less than the workfunction of the metal. Hence, no photoelectric emission will take place.
Light of frequency 7.21 × 1014 Hzis incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/sare ejected from the surface. What is the threshold frequency for photoemissionof electrons?
Answer - 10 : - Frequency of the incident photon,
Maximum speed of the electrons, v =6.0 × 105 m/s
Planck’s constant, h =6.626 × 10−34 Js
Mass of an electron, m =9.1 × 10−31 kg
For threshold frequency ν0,the relation for kinetic energy is written as:
Therefore, the threshold frequency for thephotoemission of electrons is 