Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

$X$      $Y$ because n      p + e ^- + $\stackrel{-}{\nu }$

From Fig 42.6 of the textbook, we get

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

$I$      $Pt$

From Fig 42.6 of the textbook, we can say that $Pt$ is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.