Question

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

Answer 1

What he said ⬆️

Step-by-step explanation:

Answer 2

`_(53)^(131)I \rightarrow _(54)^(131)Xe + e + \bar{\\u}`

Step-by-step 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{\\u}` (1)

where

X is the original nucleus

Y is the daughter nucleus

e is the electron

`\bar{\\u}` 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{\\u}`