Question

The isotopes of promethium, ^144₅₉ Pr and ¹³⁴₅₉ Pr are unstable, and lie on opposite sides of the “line of stability”. Which of the following combinations is most likely to represent the type of decay for these isotopes? ///// So the answer is Beta decay for promethium-144 and positron decay for promethium-134 but can someone please explain why this is the case?

N/Z's are 1.44 and 1.27 respectively... what's considered a "very high" versus a "high" versus a "low" ratio of neutrons to protons? I can't seem to find clear thresholds. Thanks!

Answer 1

Actually there is no limit over high or low value this type of problem is solved with reference of stable one which is ₅₉Pr¹⁴¹

if unstable nuclei have greater n / p ratio than stable nuclei it undergo beta decay to decrease it's n / p ( as in case of ₅₉Pr¹⁴⁴⁾ ​​where a neutron is transformed into proton

But if n / p ratio has to increase ( as in ₅₉ Pr^{​​​​​134} ) it undergo positron decay where a proton convert to neutron.