Question
Question: The \( \alpha \) (alpha) and \( \beta \) (beta) particles emitted when uranium nucleus \( _{92}{{\te...
The α (alpha) and β (beta) particles emitted when uranium nucleus 92U238 decay to 82Pb214 are:
(A) 6 α particles and 2 β particles
(B) 4 α particles and 2 β particles
(C) 2 α particles and 6 β particles
(D) 2 α particles and 4 β particles
Solution
We know that for a reaction to take place, the mass number and atomic number on the product side must be equal to that on the reactant side. So, they compared the equations for mass number and atomic number on both sides, the value of alpha and beta can be calculated.
Complete step by step solution
The given reaction is:
92U238→82Pb214+a 2He4+b −1e0
We assume that “a” number of (alpha) α particles and “b” number of β (beta) particles are emitted.
Now, we know that for every reaction, the mass number and atomic number must be conserved, that is, mass number and atomic number on the product side must be equal to that on the reactant side.
Firstly, on equating the mass numbers on both sides, we get
238=214+4 a+0238=214+4 a 4 a=24 a=424 a=6
Therefore, 6 alpha particles are emitted
Now, on equating the atomic numbers on both sides, we get:
92=82+2 a−b92−82=2 a−b92−82=2×6−b
92−82−12=−b−b=−2b=2
Therefore, 2 beta particles are emitted.
So, the correct option is (A).
Note
Alpha particles consist of 2 protons and 2 neutrons bound together into a particle identical to He4 nucleus.
They are generally produced in the proven of alpha decay, The electric charge is +2 e , mass is 6⋅6446×10−27 kg and spin=0.
Beta particles are a high energy, high speed electron or positron which is emitted by the radioactive decays of an atomic nucleus during the process of beta decay. It has charge −1 ,Mass =20001 th of a proton and spin =21