15.7k views
4 votes
The following nuclei do not lie in the band of stability. How would they be expected to decay?

a) 19/9F (Fluorine-19)
b) 39/19K (Potassium-39)
c) 55/28Ni (Nickel-55)
d) 75/33As (Arsenic-75)

User Bvanlew
by
8.1k points

1 Answer

3 votes

Final answer:

The given nuclei outside the band of stability are expected to decay to achieve a more stable state. Fluorine-19 is expected to decay by beta-minus decay, Potassium-39 and Arsenic-75 by electron capture, and Nickel-55 by beta-plus decay.

Step-by-step explanation:

The nuclei that do not lie in the band of stability are unstable and are expected to decay to achieve a more stable state. Let's discuss each of the given nuclei:

a) 19/9F (Fluorine-19): Fluorine-19 is an unstable isotope of fluorine. It is expected to decay by beta-minus decay, where a neutron inside the nucleus is converted into a proton while emitting an electron (beta particle) and an antineutrino.

b) 39/19K (Potassium-39): Potassium-39 is also an unstable isotope. It is expected to undergo electron capture, where an electron from the inner shell is captured by a proton in the nucleus, forming a neutron and emitting a neutrino.

c) 55/28Ni (Nickel-55): Nickel-55 is an unstable isotope of nickel. It is expected to decay by beta-plus decay, where a proton in the nucleus is converted into a neutron while emitting a positron (positive electron) and a neutrino.

d) 75/33As (Arsenic-75): Arsenic-75 is an unstable isotope of arsenic. It is expected to undergo electron capture, similar to potassium-39, converting a proton into a neutron and emitting a neutrino.

These are the expected decay modes for the given nuclei based on their position outside the band of stability.

User Bxdoan
by
8.3k points