Please help break down the following question in a step by step easy to comprehend process.

When one atom of iodine-131 decays by Beta particle emission, 1.5541 X 10^ -33 J are released. The atomic mass of the product nuclide is 130.9051 atomic mass units. Write a nuclear reaction for the decay of iodine-131 and determine its atomic mass in amu. 1 amu = 1.66053886 X 10^ -27kg; mass of a B particle = 9.10938291 X 10^ -31kg

Respuesta :

Answer:

The atomic mass of  iodine-131 is 130.9056485 amu.

Explanation:

Beta-decay is the process in which a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

[tex]_{53}^{131}\textrm{I}\rightarrow _{54}^{131}\textrm{Xe}+_{-1}^0\beta[/tex]

Energy released during beta particle emission = E

[tex]E=\Delta mc^2[/tex]

[tex]1.5541\times 10^{-33} J=\Delta m(3\times 10^8 m/s)^2[/tex]

[tex]\Delta m=1.7267\times 10^{-50} kg[/tex]

[tex]\Delta m=1.0399\times 10^{-23} amu[/tex]

Mass of reactant = Mass of product + Δm

=  Mass of beta particle + nuclide + Δm

Mass Beta-particle =[tex]9.10938291\times 10^{-31} kg=0.000548579 amu[/tex]

Mass of reactant =

=130.9051+0.000548579 amu +[tex]1.0399\times 10^{-23}[/tex] amu

= 130.9056485 amu

The atomic mass of  iodine-131 is 130.9056485 amu.