Question

Write the nuclear reaction equation for the beta decay of Iodine-131

Answer 1

Answer:

$_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}$

Explanation:

In a beta (minus) decay, a neutron in a nucleus turns into a proton, emitting a fast-moving electron (called beta particle) alongside with an antineutrino.

The general equation for a beta decay is:

$^A_Z X \rightarrow _{Z+1}^AY+^0_{-1}e+ ^0_0\bar{\nu}$ (1)

where

X is the original nucleus

Y is the daughter nucleus

e is the electron

$\bar{\nu}$ is the antineutrino

We observe that:

• The mass number (A), which is the sum of protons and neutrons in the nucleus, remains the same in the decay
• The atomic number (Z), which is the number of protons in the nucleus, increases by 1 unit

In this problem, the original nucles that we are considering is iodine-131, which is

$_{53}^{131}I$

where

Z = 53 (atomic number of iodine)

A = 131 (mass number)

Using the rule for the general equation (1), the dauther nucleus must have same mass number (131) and atomic number increased by 1 (54, which corresponds to Xenon, Xe), therefore the equation will be:

$_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}$

Answer 2

Answer:

What he said ⬆️

Explanation: