Is I-131 more likely to undergo beta-decay or positron-decay?

1 Answer
Jan 7, 2015

Answer:

Iodine-131 is more likely to undergo beta decay than positron decay.

Explanation:

Iodine-131 is more likely to undergo beta decay than positron decay.

Without going into too much detail, an element's radioactivity corresponds to the ratio between protons and neutrons it has in its nucleus.

If an element's neutron to proton ratio (N/Z ratio) is too high, which means that it has too many neutrons in its nucleus, the isotope is neutron rich, thus unstable. If it's the other way around, i.e. it has too many protons in its nucleus, the isotope is again unstable.

The N/Z ratio must fall within a certain range, known as the Belt of Stability, in order for the element to be stable.

http://chemwiki.ucdavis.edu/Physical_Chemistry/Nuclear_Chemistry/Nuclear_Stability_and_Magic_Numbers

Iodine-131 has a mass number of 131. Since its atomic number is 53, we can determine that is has

#131-53 = 78# #"neutrons in its nucleus"#

This means that the N/Z ratio is high (78 neutrons compared to 53 protons), which means that I-131 will undergo beta decay, since this type of decay will reduce the N/Z ratio.

The equation for I-131's beta decay is

#""_53^131I -> ""_54^131Xe + beta#, where #beta# is a beta-particle (an electron).

Notice that the number of protons has increased by 1 (53 #-># 54), but the atomic mass number has remained unchanged. There are now 54 protons and #131-54 = 77# neutrons, which means the N/Z ratio has decreased and the stability of the isotope has increased.

Here's a link to another answer a contributor already posted:

http://socratic.org/questions/can-you-write-the-nuclear-decay-equation-for-the-beta-decay-of-iodine-131-b-and-