Using the relation between the radius of nucleus and atomic mass,
Atomic mass of gold, A1 = 197
Atomic mass of silver, A2 = 107
Now, taking log on both sides,
= 1.23, which is the required ratio of the nucleii.
(i) Considering the first reaction,
Q-value is given by,
Since, Q-value is negative, this reaction is endothermic.
(ii) The second reaction is,
Q-value is given by,
Since, the Q-value is positive , the reaction is exothermic.
The nucleus decays by β- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:
The of may be represented as:
Ignoring the rest mass of antineutrino and electron , we get
Mass defect,
This energy of 4.3792 MeV, is shared by pair because, is very massive.
The maximum K.E. of when energy carried by is zero.
For the given reaction, mass defect is,
Now, Q-value is ,
which, is the maximum energy of the positron.
We have,
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 Ee≈ Q.