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Suppose, we think of fission of a Fe2656 nucleus  into two equal fragments Al1328. Is the fission energetically possible? Argue by working out Q of the process.
Given m(Fe2656) = 55.93494 u,  m (Al1328) = 27.98191 u.

Q- value is given by,

Q = [m (Fe2656) - 2m(Al1328) × 931.5 MeV     = [55.93494 - 2 × 27.98191] × 931.5 MeV 

 Q= -0.02886 × 931.5 MeV = -26.88 MeV, which is negative. 

Since, the Q-value is negative (energy released from the reaction is negative), the fission is not possible energetically.
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The radionuclide C11 decays according to C611  B511+e++v;  T1/2 = 20.3 min.
The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:
mC611 = 11.011434 u  and  m B511 = 11.009305 u.
Calculate Q and compare it with the maximum energy of the positron emitted.

For the given reaction, mass defect is, 
                               m =  [m C611 - 6me - m B511 - 5 me+me        =  mC611 - m(B511) -  2 me        =  11.011434 u - 11.009305 u - 2 × 0.000548 u        =  0.001033 u

Now, Q-value is , 

              Q = 0.001033 × 931.5 MeV     = 0.962 MeV 

which, is the maximum energy of the positron. 
We have, 
                     Q = Ed+Ee+Ev 
The daughter nucleus is too heavy compared to e+ and v. So, it carries negligible energy (Ed ≈ 0). If the kinetic energy (Ev) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q.
Hence, maximum E
e≈ Q.

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Obtain approximately the ratio of the nuclear radii of the gold isotope Au79197 and the silver isotope Ag47107.


Using the relation between the radius of nucleus and atomic mass,
                             R  A1/3 

Atomic mass of gold, A1 = 197 
Atomic mass of silver, A2 = 107

                       R1R2 = A1A21/3 

                                = 1971071/3 = (1.84)1/3 

Now, taking log on both sides, 

         log10R1R2 = log10(1.84)1/3 

         log10R1R2 = 13log10(1.84)             

                              = 13×0.2648= 0.08827 

                   R1R2 = antilog(0.08827) 
                              = 1.23, which is the required ratio of the nucleii. 

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The nucleus Ne1023 decays by β- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:
m(Ne1023) = 22.994466 um(Na1123) = 22.989770 u


The β-decay of Ne1023 may be represented as:

            Ne1023  Na1123 - e-10 + v¯ + Q 

Ignoring the rest mass of antineutrino v¯ and electron , we get 

Mass defect,                  
                   m = m(Ne1023) - m (Na1123)         = 22.994466 - 22.989770          = 0.004696 u          

  Q = 0.004696×931 MeV = 4.372 MeV. 

This energy of 4.3792 MeV, is shared by e- and v¯ pair because, Na1123 is very massive.

The maximum K.E. of e- = 4.372 MeV, when energy carried by v¯ is zero.

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The Q-value of a nuclear reaction A + b → C + d is defined by
Q = [m
A + mb - mc - md] c2
where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i) H11+H13  H12+H12 
(ii) C612+C612  Ne1020 + He24
Atomic masses are given to be
                      m(H12) = 2.014102 u;  m (H31) = 3.0160409 um (C612) = 12.000000  u; m (Ne2010)  = 19.992439 u.


(i) Considering the first reaction,

                    H11 +H13  H12 + H12 

Q-value is given by,
                     Q = m×931.5 MeV     = [m(H11) + m(H13) - 2m (H12)] × 931 MeV     = [1.007825+3.016049 -2×2.014102] × 931 Mev     = -4.03 MeV 
Since, Q-value is negative, this reaction is endothermic.

(ii) The second reaction is, 

                  C612 + C612  Ne1020 + He24 

Q-value is given by,
                      Q = m×931 MeV     = [2 m (C612) - m (Ne1020) - m(He24)] × 931 MeV     = [24.000000 - 19.992439 - 4.002603] × 931 MeV     = + 4.61 MeV

Since, the Q-value is positive , the reaction is exothermic.

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