Hydrogen-3 (3h) carbon-14 (14c) oxygen-16 (16o) which isotope (s) would most likely undergo nuclear decay, and why? hint: hydrogen's atomic number = 1; carbon's atomic number = 6; oxygen's atomic number = 8 hydrogen-3 would most likely decay because odd-numbered isotopes are less stable. carbon-14 and oxygen-16 would most likely decay because their atomic numbers are higher. carbon-14 and hydrogen-3 would most likely decay because they have low atomic mass. oxygen-16 would most likely decay because its neutrons and protons are equal. carbon-14 and hydrogen-3 would most likely decay because their neutrons and protons are not equal. oxygen-16 would most likely decay because it has the largest atomic mass.

Question

Chemistry

Maxine

4 September, 23:39

Hydrogen-3 (3h) carbon-14 (14c) oxygen-16 (16o) which isotope (s) would most likely undergo nuclear decay, and why? hint: hydrogen's atomic number = 1; carbon's atomic number = 6; oxygen's atomic number = 8 hydrogen-3 would most likely decay because odd-numbered isotopes are less stable. carbon-14 and oxygen-16 would most likely decay because their atomic numbers are higher. carbon-14 and hydrogen-3 would most likely decay because they have low atomic mass. oxygen-16 would most likely decay because its neutrons and protons are equal. carbon-14 and hydrogen-3 would most likely decay because their neutrons and protons are not equal. oxygen-16 would most likely decay because it has the largest atomic mass.

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Zachariah Jacobson · Answer 1

The correct answer will be:

carbon-14 and hydrogen-3 would most likely decay because their neutrons and protons are not equal.

Due to atoms tend to decay when it has excess of either protons or neutrons