Consider a weakly radioactive element with an atomic number 90 which tarnishes when exposed to air and
radioactive decay. At the end of the radioactive chain, it emits six alpha particles and four beta particles.
Identify the radioactive element, and also find its end nucleus atomic and mass number
Answers
Element is Thorium which has an atomic no 90 and mass no 232.
Final end nucleus will be tungsten or lead, depending on positive or negative decay, with mass no 208.
Explanation:
an alpha particle emission results in the decrease of mass by 4 and atomic no by 2, since an alpha particle is a helium atom.
Emission of beta particle leads to the decrease of one electron.
Here 6 alpha particles are emitted.
hence mass decrease= 6×4= 24
Atomic no decrease= 6×2= 12
so elemnt will be the on with atomic no 90-12=78
and mass no 232-24=208
Hence the final element will be Platinum (Pt)
when 4 Beta emission occurs,
It can positive or negative beta decay.
Postive results in decrease of one atomic no and negative decay results in increase of one atomic no.
When positive occurs atomic no=78-4=74
element is tungsten(W). Mass no doesn't change.
when negative occurs, atomic no is 78+4= 82
Element will be Lead (Pb).
Answer:
The atomic number of the new element is 82 and the mass number is 208. The element is lead.
Explanation:
An alpha particle emission results in an atom whose mass number is lower by 4 and atomic no is lower by 2 than the parent atom.
The emission of the beta particle results in an increase in the atomic number by 1; the mass number remains the same.
When 6 alpha particles are emitted, the mass decreases by 24 (6×4).
The atomic number decrease= 6×2= 12
So the atomic number becomes 78(90-12).
Given, that the mass number is 232
The new mass number becomes 232-24=208
When four Beta emission occurs, the atomic number increases by 4.
The atomic number becomes 82(78+4).
The element is lead with the atomic number of the new element is 82 and the mass number is 208.