Question

Which statement correctly identifies the nuclide that is most likely to be unstable and describes why?

a) Carbon-14 (14C) is likely an unstable nuclide because it has 8 neutrons and 6 protons, and the ideal neutron-to-proton ratio is 1:5.

b) Uranium-238 (238U) is likely an unstable nuclide because it has a neutron-to-proton ratio that is 1.6:1 and the ideal neutron-to-proton ratio is 1:1.

c) Carbon-14 (14C) is likely an unstable nuclide because it has 8 neutrons and 6 protons, and the ideal neutron-to-proton ratio is 1:2.

d) Uranium-238 (238U) is likely an unstable nuclide because it has a neutron-to-proton ratio that is 1.6:1 and the ideal neutron-to-proton ratio is 1:2.

Answer 1

the answer would be B.

U-238 has a n to p ration of 1.6:1. 146 neutrons and 92 protons.

It is actually the most commonly used isotope is reactors.

C-14 is also a radioactive isotope with 8 neutrons and 6 protons.

The usual and ideal n to p ratio is 1:1 such as C-12 or Mg-24