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

1 Answer
Feb 6, 2016

Answer:

#"Kr-85"# decays by #β^-# emission, while #"Am-241"# undergoes α decay.

Explanation:

#"n/p ratio:"#

The stability of a nucleus depends on the relative number of neutrons and protons — the neutron/proton ratio (#"n/p"#).

band of stability
(from www.kentchemistry.com)

If #"n/p"# is too low (too few neutrons), the nucleus will decay by #β^-# emission.

If #"n/p"# is too high (too many neutrons), the nucleus usually undergoes #β^+# or α decay.

#""_36^85"Kr"#:

#"Kr-85"# has 36 protons and 49 neutrons.

#n/p = 49/36 = 1.36#

The stable value is closer to 1.3, so #"Kr"-85# decays by #β^-# emission.

#""_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.

#n/p = 48/37 = 1.30#

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

#""_95^241"Pu":#

#""_95^241"Pu"# has 95 protons and 146 neutrons.

Every nucleus with more than 82 protons is radioactive.

#""_95^241"Pu"# decays by emission of an α-particle to #""_93^237"Np"#.

#""_95^241"Pu" → ""_93^237"Np" + ""_2^4"He"#.