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Atomic Structure Test - 28
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Atomic Structure Test - 28
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  • Question 1/15
    1 / -0

    What will be the wavenumber of yellow radiation having wavelength 240 nm?

    Solutions

    Wavelength of yellow light = 240 nm
    240 x 10-9 m

  • Question 2/15
    1 / -0

    What will be the energy of one photon of radiation whose frequency is 5 × 1014 Hz?

    Solutions

    The energy of one photon is given by the expression, E = hν
    where, h = 6.626 × 10−34 Js and ν= 5 × 1014 s−1
    E = 6.626 × 10−34 × 5 × 1014 = 3.313 × 10−19J

  • Question 3/15
    1 / -0

    The energy of a photon is given as 3.03 x 10-19 J/atom. The wavelength of the photon is

    Solutions


  • Question 4/15
    1 / -0

    What will be the energy of a photon which corresponds to the wavelength of 0.50 Å?

    Solutions

    E = hv = hc/λ

  • Question 5/15
    1 / -0

    Compare the energies of two radiations E1 with wavelength 800 nm and E2 with wavelength 400 nm.

    Solutions


  • Question 6/15
    1 / -0

    Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. What is the ionisation energy of sodium per atom?

    Solutions

    λ = 242 nm = 242 x 10-9 m
    Energy required to ionise one atom of Na, E = hc/λ 

  • Question 7/15
    1 / -0

    The energy difference between the ground state of an atom and its excited state is 3 x 10-19 J. What is the wavelength of the photon required for this transition?

    Solutions

    ΔE = hc/λ

  • Question 8/15
    1 / -0

    A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

    Solutions

    Power of the bulb = 100 watt = 100 J s-1 Energy of one photon E = hv = hc/λ

    Number of photons emitted 

  • Question 9/15
    1 / -0

    Mark the incorrect statement regarding the photoelectric effect.

    Solutions

    The number of electrons ejected is directly proportional to the intensity of light

  • Question 10/15
    1 / -0

    A certain metal when irradiated by light (v = 3.2 x 1016 Hz) emits photoelectrons with twice K.E. as did photoelectrons when the same metal is irradiated by light (v = 2.0 x 1016 Hz). The v0 of the metal is

    Solutions

    (K.E.)1 = hv1 - hv0
    (K.E)2 = hv2 - hv0 
    As (K.E)1 = 2 x (K.E.)2
    ∴ (hv1 - hv0)0) = 2(hv2 - hv0)
    or v0 = 2v2 - v1
    = 2 x (2 x 1016) - (3.2 x 1016
    = 0.8 x 1016 Hz or 8 x 1015 Hz

  • Question 11/15
    1 / -0

    The spectrum of white light ranging from red to violet is called a continuous spectrum because

    Solutions

    In a continuous spectrum, the colours merge into each other in a continuous pattern.

  • Question 12/15
    1 / -0

    The emission spectrum of hydrogen is found to satisfy the expression for the energy change ΔE (in joules) such that  where n1 = 1, 2, 3,.... and n2 = 2, 3, 4. The spectral lines corresponds to Paschen series if

    Solutions

    For Paschen series, n1 = 3 and n2 = 4,5, 6...

  • Question 13/15
    1 / -0

    The wavelength of visible light is

    Solutions

    Visible light wavelength is from 380 nm to 760 nm.

  • Question 14/15
    1 / -0

    Which of the following types of spectrum is best depicted by the given figure?

    Solutions

    Since the sample is an 'Excited sample', therefore this can not be absorption spectra but is an emission spectrum. Also, the light disperses from the sample towards the detector direction.

  • Question 15/15
    1 / -0

    Match the constants given in column I with their values given in column II and mark the appropriate, choice.

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