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NEET-XII-Physics

11: Dual Nature Of Radiation And Matter

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  • #
    3 - Dual Nature Of Radiation And Matter
  • Qstn #1
    Find the
    Ans : 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
  • #1-a
    maximum frequency, and
    Ans : 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 × 10-34 Js



    Hence, the maximum frequency of X-rays produced is
  • #1-b
    minimum wavelength of X-rays produced by 30 kV electrons.
    Ans : The minimum wavelength produced by the X-rays is given as:



    Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
  • Qstn #2
    The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
    Ans : Work function of caesium metal,

    Frequency of light,
  • #2-a
    maximum kinetic energy of the emitted electrons,
    Ans : The maximum kinetic energy is given by the photoelectric effect as:



    Where,

    h = Planck’s constant = 6.626 × 10-34 Js



    Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
  • #2-b
    Stopping potential, and
    Ans : For stopping potential, we can write the equation for kinetic energy as:



    Hence, the stopping potential of the material is 0.345 V.
  • #2-c
    maximum speed of the emitted photoelectrons?
    Ans : 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 × 10-31 kg



    Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
  • Qstn #3
    The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
    Ans : Photoelectric cut-off voltage, V0 = 1.5 V

    The maximum 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 given experiment is 2.4 × 10-19 J.
  • Qstn #4
    Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
    Ans : 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
  • #4-a
    Find the energy and momentum of each photon in the light beam,
    Ans : The energy of each photon is given as:



    The momentum of each photon is given as:


  • #4-b
    How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
    Ans : 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:


  • #4-c
    How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
    Ans : 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


  • Qstn #5
    The energy flux of sunlight reaching the surface of the earth is 1.388 × 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.
    Ans : Energy flux of sunlight reaching the surface of earth, Φ = 1.388 × 103 W/m2

    Hence, power of sunlight per square metre, P = 1.388 × 103 W

    Speed of light, c = 3 × 108 m/s

    Planck’s constant, h = 6.626 × 10-34 Js

    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.
  • Qstn #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 × 10-15 V s. Calculate the value of Planck’s constant.
    Ans : 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 constant is
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