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

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

Explanation:

The beta decay is given by:

Where:

A: is the mass number

Z: is the number of protons  

β⁻: is a beta particle = electron

: is an antineutrino

The neutral atom has 88 electrons, so:

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

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

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!