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Dual Nature of Matter and Radiation Test - 3

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Dual Nature of Matter and Radiation Test - 3
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Weekly Quiz Competition
  • Question 1
    1 / -0.25

    Work function of a metal is the

    Solution

    In the photoelectric effect, the work function is the minimum amount of energy (per photon) needed to eject an electron from the surface of a metal.Electrons ejected from a sodium metal surface were measured as an electric current.The minimum energy required to eject an electron from the surface is called the photoelectric work function.
    This energy (work function) is a measure of how firmly a particular metal holds its electrons. The work function is important in applications involving electron emission from metals, as in  photoelectric devices and cathode-ray tubes.

  • Question 2
    1 / -0.25

    Photon has a charge of

    Solution

    A photon is massless, has no electric charge, and is a stable particle. In vacuum, a photon has two possible polarization states. The photon is the gauge boson for electromagnetism, and therefore all other quantum numbers of the photon (such as lepton number, baryon number, and flavour quantum numbers) are zero.
     

  • Question 3
    1 / -0.25

    The work function of a photoelectric material is 3.32 eV. The threshold frequency will be equal to

    Solution

    Threshold frequency: - frequency of minimum energy required to remove electron
     
    work function=3.3ev
    f=3.3 ×1.6 ×10−19 J ​/6.626 ×10−34 J −s
    =0.8 ×1015 Hz
    =8 ×1014 Hz

  • Question 4
    1 / -0.25

    Energy of a photon of green light of wavelength 5500m  is (given: h = 6.62  ×10−34 Js−1 ) approximately

    Solution

    As we know,
    the formula for energy of photon in terms of wavelength is,, E = hc/ λ
    where, E = energy of photon
    h = Planck 's constant =
    = 6.63 ×10-34 Js
    c = speed of light
    = 3 ×108 m/s
    and lambda = wavelength
    so,
    E = [6.63 ×10-34 J. s ×3 ×108 m/s]/5500 ×10-10 m
    = [0.36 ×10-26 ]/10-8
    = 0.36 ×10-18
    = 3.6 ×10-19 J
    Or, 2.26ev
     

  • Question 5
    1 / -0.25

    Uniform electric and magnetic fields are produced pointing in the same direction. An electron is projected pointing in the same direction, then

    Solution

    Electromagnetic force on electron will be zero V ×B=0
    Electrostatic force will be Fe ​=−E backward, hence electron decelerate and velocity will decrease in magnitude.

  • Question 6
    1 / -0.25

    In which case is electron emission from a metal not known?

    Solution

    The correct answer would be option B. Applying a very strong magnetic field.
    Applying a very strong magnetic field to a metal we don ’t know the electron emission.

  • Question 7
    1 / -0.25

    Photons can be

    Solution

    A photon of wavelength 6000 nm collides with an electron at rest. After scattering, the wavelength of the scattered photon is found to change by exactly one Compton wavelength.

  • Question 8
    1 / -0.25

    Given h = 6.6  ×10−34 joule sec, the momentum of each photon in a given radiation is 3.3  ×10−29 kg metre/sec. The frequency of radiation is

    Solution

    Frequency=C/ λ
    λ=h/P
    frequency=C P/ h
    f=(3 ×108 ×3.3 ×10-29 )/6.6 ×10-34
    f=3 ×1013 /2
    f=1.5 ×1013 Hz

  • Question 9
    1 / -0.25

    If the work function of a material is 2eV, then minimum frequency of light required to emit photo-electrons is  

    Solution

    Φ= h νλ=>2eV= 6.626 x 10-34 x ν

    2 x 1.6 x 10-19 = 6.626 x 10-34 x ν

    On solving we get ν= 4.6 x 1014 Hz.

  • Question 10
    1 / -0.25

    In Thomson ’s method for finding specific charge of positive rays, the electric and magnetic fields are

    Solution

    The electric and magnetic fields are Parallel and separated because magnetic field applies force only on moving charge and electric field applies force both moving and stationery charges.

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