If potassium-40 undergoes radioactive decay by losing an electron, the daughter isotope is what?

1 Answer
Jan 12, 2016

Answer:

#""^40"Ca"#

Explanation:

First, make sure that you understand what happens when a radioactive nuclide decays by losing an electron - this is known as beta decay.

As you know, an electron, which is also known as a #beta# particle, is negatively charged.

The nucleus of an atom contains protons, which are positively charged particles, and neutrons, which are neutral particles.

When a #beta# decay takes place, a neutron is being converted to a proton, an electron, and an antineutrino, #bar(nu)_e#, the antiparticle of a neutron. The electron and the antineutrino are then emitted from the nucleus, leaving behind an extra proton.

http://www.atnf.csiro.au/outreach/education/senior/cosmicengine/sun_nuclear.html

This means that the atomic number of the isotope, which as you know tells you how many protons are found in its nucleus, will increase by #1#.

On the other hand, the mass number of the isotope will remain unchanged, since a neutron is being converted to a proton.

This means that you can write

#""_19^40"K" -> ""_20^40"X" + beta^(-) + bar(nu)_e#

A quick look in the periodic table will show that element #"X"# is calcium.

Therefore, the daughter isotope that results from the beta decay of potassium-40 is calcium-40.

#""_19^40"K" -> ""_20^40"Ca" + beta^(-) + bar(nu)_e#