Calculation of true thickness of the bed

 

19)  Nuclide A decays to nuclide B exclusively through α and β decay, such that the mass number is reduced by 32 and the atomic number is reduced by 10. The number of β particles emitted during the decay of nuclide A to B is_____

 (Special Thanks to Y.Suresh Sir, GSI)

(Thanks to Chandrasekhar, ANU)

 

Solution:

 

 A nuclide may be denoted as $^{Z}N_{A}$

   Where Z is the atomic mass number

              A is the atomic number

The given data explains us like this

 

$^{Z}N_{A}=_{A-10}N ^{Z-32}+_{2}\alpha ^{4}+_{-1}\beta ^{0}$

 

For the number of $\alpha$ particles decay

   Z=Z-32+4($\alpha$)+0($\beta$)

    Z=Z-32+4($\alpha$)+0($\beta$)

    4($\alpha$)=32

   $\alpha$=8

  $\alpha$paritcles=8

 

For the number of $\beta$ particles decay

   A=A-10+2($\alpha$)-1($\beta$)

   A=A-10+2(8)-1($\beta$)

   $\beta$=-10+16

 $\beta$=6

$\beta$particles -6