March 20, 2012

Nuclear Chemistry Three Marks

1. Calculate Q value of the following nuclear reaction
13Al27 + 2He4 14Si30 + 1H1 + Q. The exact mass of 13Al27 is 26.9815 amu, 14Si30 is 29.9738, 2He4 is 4.0026 amu and 1H1 is 1.0078 amu.
Q value of the nuclear reaction = (mp - mr) x 931 MeV
Δm = (Sum of the masses of products, mp - Sum of the masses of reactants, mr)
Δm = (29.9738 + 1.0078) - (26.9815 + 4.0026)
       = - 0.0025 amu
    Q = 0.0025 × 931 MeV
                    = 2.328 MeV

Or                              Q value of the nuclear reaction = (mr – mp) x 931 MeV
Δm = (Sum of the masses of reactants, mr – Sum of the masses of products, mp)
Δm = (26.9815 + 4.0026) – (29.9738 + 1.0078)
                  = 30.9841 – 30.9816 
                  = + 0.0025 amu
              Q = 0.0025 × 931 MeV 

                  = 2.328 MeV 
2. Calculate the decay constant for Ag108 if its half life is 2.31 minutes.
Half life, t½          = 2.31 minutes
Decay constant, λ = ?
λ   = 0.693 / t½
                 = 0.693 / 2.31
            λ = 0.3 minutes-1
3. Calculate the number of a and b particles emitted when 90Th232 nucleus is converted into 82Pb208
Let ‘a’ and ‘b’ be the number of α and β particles emitted during the change
90Th232  82Pb208 + a 2He4 + b -1e0
Comparing the mass numbers,
232 = 208 + 4a + (b × 0)
4a = 232 - 208
     = 24 / 6
                          a = 6
Comparing the atomic numbers
          90 = 82 + 2 × a + (-1)b
                           = 82 + 2a - b
    2a - b = 90 - 82 = 8
 2(6) - b = 8
                       b = 12 - 8 = 4
Number of a - particle emitted  = 6
Number of b - particles emitted = 4
4. Determine the average life of U238 having t1/2 = 140 days. Or The half-life period of U238 is 140 days. Calculate the average life time.
        Half-life, t½ = 140 days
       Average life, τ (Tau) = ?
t½ = 0.693 / λ
t½ = 0.693 x τ                             [Since, Average life, τ (Tau) = 1 / λ
              τ  = t½ / 0.693
      = 140 / 0.693
         Average life = 202.02 days
5. Explain the principle behind the 'Hydrogen bomb'.
1. Hydrogen bomb is based on the principle of nuclear fusion reactions of hydrogen to form helium producing large amount of energy.
2. Hydrogen bomb consists of an arrangement for nuclear fission in the centre surrounded by a mixture of deuterium (1H2) and lithium isotope (3Li6).
3. Fission reaction provides the high temperature necessary to start the fusion.
4. Fusion reactions take place in hydrogen bomb are
i) Fission Heat + Neutrons
ii) 3Li6 + 0n1  1H3 + 2He4 + 4.78 MeV
     1H2 + 1H3 2He4 + 0n1 + 17.6 MeV
6. Give any three differences between chemical reactions and nuclear reactions.
S.No
Chemical reactions
Nuclear reactions
1
Involve some loss, gain or overlap of outer orbital electrons of the reactant atoms.
Involve emission of alpha, beta and gamma particles from the nucleus.
2
Balanced in terms of mass only
Balanced in terms of both mass and energy.
3
Energy changes are very less when compared with nuclear reactions.
Energy changes are far exceeding
when compared with chemical reactions.
4
Energy is expressed in kJ / mol.
Energy is expressed in MeV / individual nucleus.
5
No new element is produced since nucleus is unaffected.
New element / isotope may be
produced.
7. Half-life period of 79Au198 nucleus is 150 days. Calculate its average life.
         Half-life t½ = 150 days
       Average life, τ (Tau) = ?
t½ = 0.693 / λ
t½ = 0.693 x τ                            [ Since, Average life, τ (Tau) = 1 / λ
τ  = t½ / 0.693
    = 150 / 0.693
      Average life = 216.45 days
8. How many α and β   particles will be emitted by an element 84A218 is changing to a stable isotope of 82B206?
Let ‘a’ and ‘b’ be the number of α and β particles emitted during the change
82A218  84B206 + a 2He4 + b -1e0
Comparing the mass numbers,
218 = 206 + 4a + (b × 0)
   4a = 218 - 206
                                = 12 / 4
                             a = 3
Comparing the atomic numbers
                        84 = 82 + 2 × a + (-1)b
                             = 82 + 2a - b
                   2a - b = 84 - 82 = 2
                 2(3) - b = 2
                           b = 6 - 2 = 4
Number of a - particle emitted = 3
Number of b - particles emitted = 4
9. In the conversion of 92U238  82Pb206. Calculate the number of alpha and beta particles emitted.
Let ‘a’ and ‘b’ be the number of α and β particles emitted during the change
92U238  82Pb206+ a 2He4 + b -1e0
Comparing the mass numbers,
238 = 206 + 4a + (b × 0)
  4a = 238 - 206
      = 32 / 4
   a = 8
Comparing the atomic numbers
            92 = 82 + 2 × a + (-1)b
     = 82 + 2a - b
                   2a - b = 92 - 82 = 10
    2(8) - b = 10
              b = 16 - 10 = 6
Number of a - particle emitted  = 8
Number of b - particles emitted = 6
10. In the following radioactive decay: 92X232  89Y220 how many α and β particles are ejected?
Let ‘a’ and ‘b’ be the number of α and β particles emitted during the change
92X232  89Y220 + a 2He4 + b -1e0
Comparing the mass numbers,
232 = 220 + 4a + (b × 0)
  4a = 232 - 220
                              = 12 / 4
                           a = 3
Comparing the atomic numbers,
                        92 = 89 + 2 × a + (-1)b
                             = 89 + 2a - b
                   2a - b = 92 - 89 = 3
                2(3) - b = 3
                          b = 6 - 3 = 3
Number of a - particle emitted  = 3
Number of b - particles emitted = 3
11. Neutron bombardment fragmentation of U-235 occurs according to the equation:
92U235 + 0n1  42Mo98 + 54Xe136 + x -1e0 + y 0n1. Calculate the values of x and y.
92U235 + 0n1  42Mo98 + 54Xe136 + x -1e0 + y 0n1
Comparing the mass numbers,
     235+ 1 = 98 + 136 + (x × 0) + (y × 1)
         236 = 234 + 0 + y
y = 236 - 234
  y = 2
Comparing the atomic numbers,
     92 + 0 = 42 + 54 + (-1)x + (y × 0)
           92 = 96 - x + 0
92 = 96 - x
                          x = 96 - 92
              x = 4
Number of x - particle emitted  = 4
Number of y - particles emitted = 2
12. The atomic masses of Li, He and Proton are 7.01823 amu. 4.00387 amu and 1.00715 amu respectively. Calculate the energy evolved in the reaction.
3Li7+ 1H1 → 2 2He4 + Energy (1 amu = 931MeV).
Mass of reactants                     = Mass of Li + Mass of H
= 7.01823 + 1.00715
= 8.02538 amu
Mass of products                      = 2 x Mass of He
= 2 x 4.00 387
= 8.00774 amu
Mass loss during change           = (8.02538 - 8.00774) amu
= 0.01764 amu
Energy evolved during reaction = 0.01764 x 931 MeV
= 16.423 MeV

Or

Q value = (mp– mr) x 931 MeV

                                      = (8.02538 – 8.00774) x 931 MeV
                                      = – 16.423 MeV
13. The decay constant for 6C14 is 2.31 x 10-4 year-1. Calculate the half life period.
Decay constant = 2.31 x 10-4 year-1
    Half-life, t½ = ?
        t½ = 0.693 / λ
        t½ = 0.693 / 2.31 x 10-4
             = 3000 years.
14. The half-life period of a radioactive element is 100 seconds. Calculate the disintegration constant.
                   Half-life t½ = 100 sec
Decay constant = ?
   t½ = 0.693 / λ
                 λ = 0.693 / t½
                    = 0.693 / 100
                    = 0.00693 sec-1
15. What is the Q value of a nuclear reaction?
The amount of energy absorbed or released during nuclear reaction is called Q-value of nuclear reaction.
Q value = (mp– mr) x 931 MeV
Where,
            mr - Sum of the masses of reactants.
mp - Sum of the masses of products.
Case – 1
Energy absorbed during nuclear reaction, mp > mr, and hence Q value will be positive.
Case – 2
Energy released during nuclear reaction, mr > mp, and hence Q value will be negative.
16. Write the uses of Radio carbon dating.
1. For correlating facts of historical importance.
2. In understanding the evolution of life and
3. In understanding the rise and fall of civilizations.

17. What is binding energy of nucleus?
Whenever a nucleus is formed, certain mass is converted into energy. Hence for atom, the atomic mass is lower than the sum of masses of protons, neutrons and electrons present. The difference in mass is termed as “Mass defect”. This is the measure of the Binding energy of proton and neutron in the nucleus. The relationship between mass - energy is explained by Einstein equation ΔE = Δm C2
18. What is spallation reaction? Give an example
The reactions in which high speed projectiles may chip a heavy nucleus into several fragments.
29Cu63 + 2He4 + 400 MeV → 17Cl37 + 14 1H1 + 16 0n1
19. Write the nuclear reation taking place in the sun.
It has been estimated that the sun is giving out energy equally in all possible directions at the rate of 3.7 x 1033 ergs / sec. The energy of the sun is supposed to arise from the fusion of hydrogen nuclei into helium nuclei which in going on inside it all the time.
The various fusion reactions taking place in the sun are as follows:
Proton - proton chain reaction:
                 Fusion
1H1 + 1H1 1H2 + +1e0 + Energy
                                    Positron
                  Fusion
1H2 + 1H1 2He3 + Energy
                 Fusion
2He3 +1H1 2He4 + +1e0 + Energy
The overall reaction, therefore, may be written as:
         Fusion
41H1 2He4 + 2 +1e0 + Energy


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