Question

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

Answer 1

The atomic mass of iodine-131 is 130.9056485 amu.

Step-by-step 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.

`_Z^A\textrm{X}\rightarrow _(Z+1)^A\textrm{Y}+_(-1)^0\beta`

`_(53)^(131)\textrm{I}\rightarrow _(54)^(131)\textrm{Xe}+_(-1)^0\beta`

Energy released during beta particle emission = E

`E=\Delta mc^2`

`1.5541* 10^(-33) J=\Delta m(3* 10^8 m/s)^2`

`\Delta m=1.7267* 10^(-50) kg`

`\Delta m=1.0399* 10^(-23) amu`

Mass of reactant = Mass of product + Δm

= Mass of beta particle + nuclide + Δm

Mass Beta-particle =
`9.10938291* 10^(-31) kg=0.000548579 amu`

Mass of reactant =

=130.9051+0.000548579 amu +
`1.0399* 10^(-23)` amu

= 130.9056485 amu

The atomic mass of iodine-131 is 130.9056485 amu.