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Nuclei Test - 36

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Nuclei Test - 36
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  • Question 1
    1 / -0
    23Na^{23}Na is the more stable isotope of NaNa. By which process 1124Na^{24}_{11}Na can undergo radioactive decay?
    Solution
    To convert 1124Na^{24}_{11}Na to 1123Na^{23}_{11} Na, we need to decrease the mass number of 1124Na^{24}_{11}Na by 11.
    Since, in β\beta^{-} emission and β+\beta^{+} emission, the mass number of product nucleus is same as that of parent nucleus.
    Whereas, in α\alpha emission, the mass number of product nucleus gets decreased by 44 compared to that of parent nucleus.
    Thus 1124Na^{24}_{11} Na cannot undergo any of the radioactive decay to convert into 1123Na^{23}_{11}Na.
  • Question 2
    1 / -0

    Directions For Questions

    Some of the heavy nuclei decay into lighter nuclei by emitting a-particle and acquire stability. The principles of conservation of atomic number, mass number, linear momentum and energy (including rest mass energy) are valid for these decays.
    A stationary nucleus 88226Ra^{226}_{88}Ra (ground state) decays into the nucleus 86222Rn^{222}_{86}Rn (ground state) by emitting an αparticle\alpha - particle.
    Given that m(88226Ra)=226.02540um(^{226}_{88}Ra)=226.02540 u. m(24He)=4.00260um(^4_2He)=4.00260 u. m(86222Rn)=222.01750um(^{222}_{86}Rn)=222.01750 u.

    ...view full instructions

    The K.E. of the emitted αparticle\alpha-particle in the decay of 88226Ra^{226}_{88}Ra (approximately)
    Solution
    88226Ra24He2++86222Rn \begin{array}{c} _{88}^{226} Ra \longrightarrow{ }_{2}^{4} He^{2+}+{ }_{86}^{222} Rn \end{array}  
    226.02540μ4.00260μ222.01750μ 226.02540 \mu \quad 4.00260 \mu \quad 222.01750 \mu
     Δm=0.0053 Energy =0.0053×931.5MeV=4.94K1=m2m1+m2×K=4222+4×4.94=4.8495 \begin{aligned} \Delta m &=0.0053 \\ \text { Energy } &=0.0053 \times 931.5 \mathrm{MeV} \\ &=4.94 \\ K_{1} &=\frac{m_{2}}{m_{1}+m_{2}} \times K \\ &=\frac{4}{222+4} \times 4.94 \\ &=4.8495 \end{aligned}
  • Question 3
    1 / -0
    In the nuclear reaction
    $$^{14}_7N\, +\, _2^4He\, \rightarrow\, ^{17}_8O\, +\, ...............$$
    What is the missing particle ?
    Solution
    The hydrogen isotope is missing with atomic number 1 and mass number 1.
  • Question 4
    1 / -0
    Fusion reaction is also known as ..........
    Solution
    In nuclear physics,nuclear fusion is a nuclear reaction in which two or more atomic nuclei collide at very high speed and join to form a new type of atomic nucleus. During this process matter is not conserved because some of the matter of fusing nuclei is converted to photons also known as thermonuclear reaction.
  • Question 5
    1 / -0
    Pick out the isoelectronic species from the following :
    (I) CH3+  (II)H3O+  (III)NH3  (IV)CH3(I)\, {CH_{3}}^{+}\, \quad\, (II)\, H_3O^+\, \quad\, (III)\, NH_3\, \quad\, (IV)\, {CH_{3}}^{-}
    Solution
    Isoelectronic species means the no of electron in both molecule is same.
    No. of electron in (I) is 9, in (II)=10, (III)=10, (IV)=10.
    So II, III, IV are isoelectronic species.
  • Question 6
    1 / -0
    When a hydrogen bomb explodes, which of the following is used?
    Solution
    Hydrogen bomb is based on nuclear fusion and hence hydrogen atoms combine to form Helium atoms.
  • Question 7
    1 / -0
    Two atoms of the same element are found to have different number of neutrons in their nuclei. These two atoms are :
    Solution
    (See all the isotopes of carbon have different no of neutrons.)
    Isotopes are atoms with the same number of protons but that have a different number of neutrons. Since the atomic number is equal to the number of protons and the atomic mass is the sum of protons and neutrons, we can also say that isotopes are elements with the same atomic number but different mass numbers.

  • Question 8
    1 / -0
    The isotopes present in naturally occurring uranium are

    (i)(i) U235U-235
    (ii)(ii) U236U-236
    (iii)(iii) U238U-238
    (iv)(iv) U232U-232
    Solution
    U-235 and U-238 occurs naturally.
  • Question 9
    1 / -0
    Which atom contains exactly 15 neutrons?
    Solution
    mass number=no.of protons+no.of neutronsmass\space number= no. of \space protons + no. of\space neutrons
    Given that mass number of P = 32 
    Atomic number (no. of electrons == no. of protons) == 17
    Number of neutrons =3217=15= 32 - 17 = 15
    Hence, P satisfies the requirement. 
  • Question 10
    1 / -0
    Calculate the binding energy per nucleon for Beryllium 4Be9_{4}{Be}^{9}, its mass being 9.012 u9.012 \ u. The masses of proton and neutron are 1.008 u1.008  u and 1.009 u1.009  u. (Take 1 u=931.5 MeV1  u = 931.5  MeV)
    Solution
    First, let us find the mass defect. The Beryllium nucleus contains 4 protons and 5 neutrons.
    Mass of 4 protons =4×1.008=4.032 u= 4 \times 1.008 = 4.032  u;  Mass of 5 neutrons =5×1.009 u=5.045 u= 5 \times 1.009  u = 5.045  u
    Total mass of protons (4) and neutrons (5) =4.032+5.045=9.077 u= 4.032 + 5.045 = 9.077  u
    Mass defect =9.0779.012=0.065 u= 9.077 - 9.012 = 0.065  u
    The mass defect converted into equivalent energy gives binding energy.
    1 u=931.5 MeV1  u = 931.5  MeV
     0.065 u=0.065×931.5 MeV=60.5475 MeV\therefore   0.065  u = 0.065 \times 931.5  MeV = 60.5475  MeV
    Binding energy per nucleon = Binding energyNumber of nucleons= 60.54759=6.7275 MeV= \displaystyle\frac{Binding   energy}{Number   of   nucleons} = \displaystyle\frac{60.5475}{9} = 6.7275  MeV
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