HomeLearn Fast select Course
Change UnitChange SubjectLast Menu
NEET-XII-Physics

13: Nuclei

Note: Please signup/signin free to get personalized experience.

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

  
  • #
    5 - Nuclei
  • #1
  • #1-a
    Two stable isotopes of lithium and have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.
    Ans : Mass of lithium isotope , m1 = 6.01512 u

    Mass of lithium isotope , m2 = 7.01600 u

    Abundance of , η1= 7.5%

    Abundance of , η2= 92.5%

    The atomic mass of lithium atom is given as:


  • #1-b
    Boron has two stable isotopes, and. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of and .
    Ans : Mass of boron isotope , m1 = 10.01294 u

    Mass of boron isotope , m2 = 11.00931 u

    Abundance of , η1 = x%

    Abundance of , η2= (100 - x)%

    Atomic mass of boron, m = 10.811 u

    The atomic mass of boron atom is given as:



    And 100 - x = 80.11%

    Hence, the abundance of is 19.89% and that of is 80.11%.
  • Qstn #2
    The three stable isotopes of neon: and

    have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
    Ans : Atomic mass of , m1= 19.99 u

    Abundance of , η1 = 90.51%

    Atomic mass of , m2 = 20.99 u

    Abundance of , η2 = 0.27%

    Atomic mass of , m3 = 21.99 u

    Abundance of , η3 = 9.22%

    The average atomic mass of neon is given as:


  • Qstn #3
    Obtain the binding energy (in MeV) of a nitrogen nucleus, given =14.00307 u
    Ans : Atomic mass of nitrogen, m = 14.00307 u

    A nucleus of nitrogen contains 7 protons and 7 neutrons.

    Hence, the mass defect of this nucleus, Δm = 7mH + 7mn - m

    Where,

    Mass of a proton, mH = 1.007825 u

    Mass of a neutron, mn= 1.008665 u

    ∴Δm = 7 × 1.007825 + 7 × 1.008665 - 14.00307

    = 7.054775 + 7.06055 - 14.00307

    = 0.11236 u

    But 1 u = 931.5 MeV/c2

    ∴Δm = 0.11236 × 931.5 MeV/c2

    Hence, the binding energy of the nucleus is given as:

    Eb = Δmc2

    Where,

    c = Speed of light

    ∴Eb = 0.11236 × 931.5

    = 104.66334 MeV

    Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.
  • Qstn #4
    Obtain the binding energy of the nuclei and in units of MeV from the following data:

    = 55.934939 u = 208.980388 u
    Ans : Atomic mass of, m1 = 55.934939 u

    nucleus has 26 protons and (56 - 26) = 30 neutrons

    Hence, the mass defect of the nucleus, Δm = 26 × mH + 30 × mn - m1

    Where,

    Mass of a proton, mH = 1.007825 u

    Mass of a neutron, mn = 1.008665 u

    ∴Δm = 26 × 1.007825 + 30 × 1.008665 - 55.934939

    = 26.20345 + 30.25995 - 55.934939

    = 0.528461 u

    But 1 u = 931.5 MeV/c2

    ∴Δm = 0.528461 × 931.5 MeV/c2

    The binding energy of this nucleus is given as:

    Eb1 = Δmc2

    Where,

    c = Speed of light

    ∴Eb1 = 0.528461 × 931.5

    = 492.26 MeV

    Average binding energy per nucleon

    Atomic mass of, m2 = 208.980388 u

    nucleus has 83 protons and (209 - 83) 126 neutrons.

    Hence, the mass defect of this nucleus is given as:

    Δm‘ = 83 × mH + 126 × mn - m2

    Where,

    Mass of a proton, mH = 1.007825 u

    Mass of a neutron, mn = 1.008665 u

    ∴Δm‘ = 83 × 1.007825 + 126 × 1.008665 - 208.980388

    = 83.649475 + 127.091790 - 208.980388

    = 1.760877 u

    But 1 u = 931.5 MeV/c2

    ∴Δm‘ = 1.760877 × 931.5 MeV/c2

    Hence, the binding energy of this nucleus is given as:

    Eb2 = Δm‘c2

    = 1.760877 × 931.5

    = 1640.26 MeV

    Average bindingenergy per nucleon =
  • Qstn #5
    A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of atoms (of mass 62.92960 u).
    Ans : Mass of a copper coin, m’ = 3 g

    Atomic mass of atom, m = 62.92960 u

    The total number of atoms in the coin

    Where,

    NA = Avogadro’s number = 6.023 × 1023 atoms /g

    Mass number = 63 g



    nucleus has 29 protons and (63 - 29) 34 neutrons

    ∴Mass defect of this nucleus, Δm‘ = 29 × mH + 34 × mn - m

    Where,

    Mass of a proton, mH = 1.007825 u

    Mass of a neutron, mn = 1.008665 u

    ∴Δm‘ = 29 × 1.007825 + 34 × 1.008665 - 62.9296

    = 0.591935 u

    Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022

    = 1.69766958 × 1022 u

    But 1 u = 931.5 MeV/c2

    ∴Δm = 1.69766958 × 1022 × 931.5 MeV/c2

    Hence, the binding energy of the nuclei of the coin is given as:

    Eb= Δmc2

    = 1.69766958 × 1022 × 931.5

    = 1.581 × 1025 MeV

    But 1 MeV = 1.6 × 10-13 J

    Eb = 1.581 × 1025 × 1.6 × 10-13

    = 2.5296 × 1012 J

    This much energy is required to separate all the neutrons and protons from the given coin.
  • Qstn #6
    Write nuclear reaction equations for
    (i) α-decay of (ii) α-decay of
    (iii) β--decay of (iv) β--decay of
    (v) β+-decay of (vi) β+-decay of
    (vii) Electron capture of
    Ans : α is a nucleus of helium and β is an electron (e- for β- and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β--decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

    For the given cases, the various nuclear reactions can be written as:


  • Qstn #7
    A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to
    a) 3.125%, b) 1% of its original value?
    Ans : Half-life of the radioactive isotope = T years

    Original amount of the radioactive isotope = N0

    a) After decay, the amount of the radioactive isotope = N

    It is given that only 3.125% of N0 remains after decay. Hence, we can write:



    Where,

    λ = Decay constant

    t = Time



    Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

    b) After decay, the amount of the radioactive isotope = N

    It is given that only 1% of N0 remains after decay. Hence, we can write:



    Since, λ = 0.693/T



    Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.
  • Qstn #8
    The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive present with the stable carbon isotope . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
    Ans : Decay rate of living carbon-containing matter, R = 15 decay/min

    Let N be the number of radioactive atoms present in a normal carbon- containing matter.

    Half life of, = 5730 years

    The decay rate of the specimen obtained from the Mohenjodaro site:

    R‘ = 9 decays/min

    Let N’ be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

    Therefore, we can relate the decay constant, λand time, t as:



    Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.
  • Qstn #9
    Obtain the amount of necessary to provide a radioactive source of 8.0 mCi strength. The half-life of is 5.3 years.
    Ans : The strength of the radioactive source is given as:



    Where,

    N = Required number of atoms

    Half-life of, = 5.3 years

    = 5.3 × 365 × 24 × 60 × 60

    = 1.67 × 108 s

    For decay constant λ, we have the rate of decay as:



    Where, λ





    For:

    Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

    ∴Mass of atoms

    Hence, the amount of necessary for the purpose is 7.106 × 10-6 g.
  • Qstn #10
    The half-life of is 28 years. What is the disintegration rate of 15 mg of this isotope?
    Ans : Half life of , = 28 years

    = 28 × 365 × 24 × 60 × 60

    = 8.83 × 108 s

    Mass of the isotope, m = 15 mg

    90 g of atom contains 6.023 × 1023 (Avogadro’s number) atoms.

    Therefore, 15 mg of contains:



    Rate of disintegration,

    Where,

    λ = Decay constant



    Hence, the disintegration rate of 15 mg of the given isotope is 7.878 × 1010 atoms/s.
  • Qstn #11
    Obtain approximately the ratio of the nuclear radii of the gold isotope and the silver isotope.
    Ans : Nuclear radius of the gold isotope = RAu

    Nuclear radius of the silver isotope = RAg

    Mass number of gold, AAu = 197

    Mass number of silver, AAg = 107

    The ratio of the radii of the two nuclei is related with their mass numbers as:



    Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.
  • Qstn #12
    Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of
next page