Question #2d7b6

1 Answer
Sep 30, 2015

#beta^-# decay


So, you know that chromium-51 decays into manganese-51, but you don't know what type of radiation is emitted when this takes place.

For an isotope, the number that follows the element's name represents its mass number, which tells you how many protons and neutrons the isotope has in its nucleus.

In your case, you know that you're dealing with two isotopes that have the same mass number, #51#.

However, since you're essentially dealing with the isotopes of the different elements, chromium and manganese, then you know for a fact that their atomic numbers will be different.

A quick look in the periodic table will show you that chromium's atomic number is equal to #24# and manganese's atomic number is equal to #25#.

This means that you can write

#"chromium-51" = ""_24^51"Cr"#

#"manganese-51" = ""_25^51"Mn"#

The nuclear equation for this decay will look like this

#""_24^51"Cr" -> ""_25^51"Mn" + color(blue)(?)#

So, what must happen in order for this reaction to take place?

Notice that the product of the decay has one extra proton than chromium-51, but that their mass numbers are equal.

This means that a neutron from chromium-51's nucleus was converted into a proton and an electron.

What would this imply?

Well, if a neutron is converted into a proton, the mass number will remain unchanged, but the atomic number will increase by #1#. In addition to this, in order for the charge to be preserved as well, an electron antineutrino, #bar(nu)_e#.

So, the nuclear equation will look like this

#""_24^51"Cr" -> ""_25^51"Mn" + ""_text(-1)^(0)e + bar(nu)_e#

The radioactive decay that in which an electron and an electron antineutrino are emitted from the nucleus of a radioactive isotope is called beta-minus decay, #beta^-# decay.

SIDE NOTE Actually, chromium-51 decays via electron capture to vanadium-51, so I assume that this was just a practice problem for understanding how to balance nuclear equations and how various types of radioactive decay work.