Which element results if 2 protons and 2 neutrons are ejected from a radium nucleus with atomic number 88 and mass number 226 - show workings?

What is actually ejected is an alpha particle (helium nucleus). The new atomic number is 88-2 = 86 and the new atomic weight is 226 - 4 = 222. Use a periodic table or elements list to identify the remaining nucleus.
Hint: it is a gas

thank you - very much appreciated.

RN Radon 222 / 86

To find out which element is formed when 2 protons and 2 neutrons are ejected from a radium nucleus with atomic number 88 and mass number 226, you need to consider the changes in atomic number and atomic weight.

1. Start with the original atom, which is radium (Ra) with atomic number 88 and mass number 226.
2. Two protons are ejected, so the new atomic number is 88 - 2 = 86.
3. Two neutrons are also ejected, so the new atomic weight is 226 - 4 = 222.
4. Use a periodic table or elements list to find the element with atomic number 86. The element is radon (Rn).
5. Radon has an atomic weight of 222, which matches the new atomic weight from the calculation.

Therefore, the element formed when 2 protons and 2 neutrons are ejected from a radium nucleus is radon (Rn) with atomic number 86 and atomic weight 222. Radon is a gas.

You're welcome! I'm glad I could help. If you have any more questions, feel free to ask.