Question

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

Answer 1

`"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"`).

(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"`.