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
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.
Given, mass of the coin = 3.0 g
Mass of atom = 62.92960 u
Each atom of the copper contains 29 protons and 34 neutrons.
Mass of 29 electrons = 29 x 0.000548 u
= 0.015892 u
Mass of nucleus = (62.92960 - 0.015892) u
= 62.913708 u
Mass of 29 protons = 29 x 1.007825 u
= 29.226925 u
Mass of 34 neutrons = 34 x 1.008665 u
= 34.29461 u
Total mass of protons and neutrons = (29.226925 + 34.29461) u
= 63.521535 u
Binding energy = (63.521535 - 62.913708) x 931.5 MeV
= 0.607827 x 931.5 MeV
Required energy =
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.
Half- life of = 28 years.
Using the formula,
90 g of Sr contains 6.023 x 1023 atoms.
15 mg of Sr contains,
For the given reaction, mass defect is,
Now, Q-value is ,
which, is the maximum energy of the positron.
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.