# How is the radioactive decay of krypton-85 different from the radioactive decay of Americium-241?

Feb 6, 2016

$\text{Kr-85}$ decays by β^- emission, while $\text{Am-241}$ undergoes α decay.

#### Explanation:

$\text{n/p ratio:}$

The stability of a nucleus depends on the relative number of neutrons and protons — the neutron/proton ratio ($\text{n/p}$).

(from www.kentchemistry.com)

If $\text{n/p}$ is too low (too few neutrons), the nucleus will decay by β^- emission.

If $\text{n/p}$ is too high (too many neutrons), the nucleus usually undergoes β^+ or α decay.

$\text{_36^85"Kr}$:

$\text{Kr-85}$ has 36 protons and 49 neutrons.

$\frac{n}{p} = \frac{49}{36} = 1.36$

The stable value is closer to 1.3, so $\text{Kr} - 85$ decays by β^- emission.

$\text{_36^85"Kr" → ""_37^85"Kr" + color(white)(l)_text(-1)^0"e}$

The atomic number increases by 1 unit, but the mass number stays the same.

There are now 37 protons and 48 neutrons.

$\frac{n}{p} = \frac{48}{37} = 1.30$

The product is the stable, non-radioactive nuclide, rubidium-85.

$\text{_95^241"Pu} :$

$\text{_95^241"Pu}$ has 95 protons and 146 neutrons.

Every nucleus with more than 82 protons is radioactive.

$\text{_95^241"Pu}$ decays by emission of an α-particle to $\text{_93^237"Np}$.

$\text{_95^241"Pu" → ""_93^237"Np" + ""_2^4"He}$.