Question

If a neutral atom with 88 electrons undergoes 2 rounds of beta minus decay, what will be the new element?

Answer 1

Answer:

The new element will be thorium-226 (²²⁶Th).

Explanation:

The beta decay is given by:

$^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + \beta^{-} + \bar{\nu_{e}}$

Where:

A: is the mass number

Z: is the number of protons  

β⁻: is a beta particle = electron

$\bar{\nu_{e}}$: is an antineutrino

The neutral atom has 88 electrons, so:

$e^{-} = 88 = Z$

Hence the element is radium (Ra), it has A = 226.

If Ra undergoes 2 rounds of beta minus decay, we have:

$^{226}_{88}Ra \rightarrow ^{226}_{89}Ac + \beta^{-} + \bar{\nu_{e}}$    

$^{226}_{89}Ac \rightarrow ^{226}_{90}Th + \beta^{-} + \bar{\nu_{e}}$

Therefore, if a neutral atom with 88 electrons undergoes 2 rounds of beta minus decay the new element will be thorium-226 (²²⁶Th).

I hope it helps you!