What is the nuclear decay equation for americium-244?

Answer 1

• Americium is a transuranic radioactive chemical element that has the symbol Am and atomic number 95.
• Its most common isotopes are 241Am and 243Am.
• 241Am decays by emission of an Alpha particle to 237Np; the half-life of this decay is 432.2 years. 241Am95 (by Alpha decay) ------ 237Np93
• Refer to link below for more information.

Answer 2

It gives out mainly alpha radiation: this is strongly ionising, but can be stopped by a few centimetres of air. This means the americium can do its job in the chamber, but no radiation will escape outside where it could be a health hazard. Americium does give off a small proportion of gamma rays, but these are minimal, and far below your average dose from background radiation.

Answer 3

Americium-241 decays into neptunium-237 through alpha decay. It has a half-life of 7,370 years. In which the total decay series ends in Bismouth 209

Answer 4

The equation for the beta decay of 244Am is:

95244Am --> 96244Cm + -10e

with -10e representing the beta particle or electron.

Answer 5

Am-241,95 ------------> He-4,2 (alpha particle) + Np-237,93

Answer 6

The three observed isotopes for americium decay into neptunium isotopes (Np-237, 238, or 239).

Answer 7

We know that americium-243 might undergo spontaneous fission, but it decays by alpha emission and not via beta decay.

Answer 8

Americium emits alphas, gammas and X-rays.

Answer 9

See the link below for a table of americium isotopes and their decay modes.

Answer 10

Americium-243 hasn't a beta decay (only alpha).