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Modern Physics Test - 6

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Modern Physics Test - 6
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  • Question 1
    4 / -1

    Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube?

    Solution

    The continuous spectrum depends on the accelerating voltage. It has a definite minimum wavelength.
    Greater the accelerating voltage for electrons, higher will be the kinetic energy it attains before striking the target, higher will be the frequency of X - rays and smaller will be the wavelength. The wavelength of continuous X - rays is independent of the atomic number of target material.

  • Question 2
    4 / -1

    A radioactive sample S1 having an activity 5µCi has twice the number of nuclei as another sample S2 which has an activity of 10 µCi. The half lives of S1 and S2 can be

    Solution

  • Question 3
    4 / -1

    Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions φp = 2.0 eV, φq = 2.5 eV and φr = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350  nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is  [Take hc = 1240 eV nm]

    Solution

    The energy possessed by photons of wavelen gth 550 nm is 1240/550 = 2.25eV
    The energy possessed by photons of wavelength 450 nm is 1240/450 = 2.76eV

    The energy possessed by photons of wavelength 350 nm is 1240/350 = 3.5.eV

    For metal plate p :

    φp = 2 eV.

    Photons of wavelength 550 nm will not be able to eject electrons and therefore, the current is smaller than p.
    The work function is greater than q therefore the stopping potential is lower in comparison to p.

    For metal plate r :

    φr = 3 eV

    Only wavelength of 350 nm will be able to eject electrons and therefore, current is minimum. Also the stopping potential is least.

  • Question 4
    4 / -1

    The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A°. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

    Solution

    The wave length of first spectral line in the Balmer series of hydrogen atom is 6561Å . Here n2 = 3 and n1 = 2

    For the second spectral line in the Balmer series of singly ionised helium ion n2 = 4 and n1 = 2 ; Z = 2

    Dividing equation (i) and equation (ii) we get

  • Question 5
    4 / -1

    A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3×108 ms–1. The final momentum of the object is

    Solution

  • Question 6
    4 / -1

    If λCu is the wavelength of Kα X-ray line of copper (atomic number 29) and λMo is the wavelength of the Kα X-ray line of molybdenum (atomic number 42), then the ratio λCuMo is close to

    Solution

  • Question 7
    4 / -1

    A metal surface is illuminat ed by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly

    Solution

    Dividing the above two equations, we get

  • Question 8
    4 / -1

    In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ) of incident light and the corresponding stopping potential (V0) are given below :

    Given that c = 3 × 108m s–1 and e = 1.6 × 10–19 C , Planck's constant (in units of J s) found from such an experiment is

    Solution

    From the first two values given in data

    Similarly if we calculate h  for the last two values of data  h = 6.4 × 10–34J-s

  • Question 9
    4 / -1

    The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by 

    The measured masses of the neutron  are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the  nuclei are same, 1 u = 931.5 Me V/c2 (c is the speed of light) and e2/(4 πε0) = 1.44 MeV fm. Assuming  that the difference between the binding energies of   is purely due to the electrostatic energy, the radius of either of the nuclei is (1 fm = 10–15 m)

    Solution

    Binding energy of nitrogen atom
    = [8 × 1.008665 + 7 × 1.007825 – 15.000109] × 931
    Binding energy of oxygen atom
    = [7 × 1.008665 + 8 × 1.007825 – 15.003065] × 931
    ∴ Difference = 0.0037960 × 931 MeV    ...(I)


    From (i) & (ii)

  • Question 10
    4 / -1

    An accident in a nuclear labora tory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which  the laboratory can be considered safe for use?

    Solution

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