Balance the nuclear reaction equation for the beta minus decay of iodine-131

Balance the nuclear reaction equation for the beta minus decay of iodine-131 by completing the missing numbers. es001-1.jpg The superscript of xenon is . The subscript of the electron is .

2 months ago

Solution 1

Guest Guest #1589
2 months ago

Balanced nuclear reaction equation for the beta minus decay of iodine-131 is ${}_{{\rm{53}}}^{{\rm{131}}}{\rm{I}} \to {}_{54}^{{\rm{131}}}{\rm{Xe + }}{}_{{\rm{ - 1}}}^0{\rm{e}}$.

What is nuclear reaction?

Nuclear reaction is a reaction which changes the identity of an atomic nuclei, when high energetic particles is bombarded on that.

Beta minus decay is one of the example of nuclear reaction, and in this reaction a neutron decays into an electron which is also known as beta particle having no mass in it. And at the same time gamma rays also. In the question, it is given that iodine-131 shows beta minus decay, where the superscript of xenon is 131 and subscript of electron is -1. i.e.

${}_{{\rm{53}}}^{{\rm{131}}}{\rm{I}} \to {}_{\rm{x}}^{{\rm{131}}}{\rm{Xe + }}{}_{{\rm{ - 1}}}^{\rm{y}}{\rm{e}}$

We have to balance the above equation and for that sum of the superscripts and subscripts is always equal on both left and right sides.

Calculation for superscripts:

131 = 131 + y

y = 0.

Calculation for subscripts:

53 = x - 1

x = 53 + 1 = 54

Hence, balanced nuclear equation can be written as ${}_{{\rm{53}}}^{{\rm{131}}}{\rm{I}} \to {}_{54}^{{\rm{131}}}{\rm{Xe + }}{}_{{\rm{ - 1}}}^0{\rm{e}}$.

To learn more about nuclear reactions, visit below link:

Solution 2

Guest Guest #1590
2 months ago

131            131            0
      I    →         Xe   +     e
  53             54           - 1


1) β decay is the radiactive process in which an element (Iodine - 131) in this case undergoes a desintegration by emitting beta radiation.

2) Beta radiation are electrons.

3) The electron has a negative charge 1-, and no mass (almost  not mass), so it is represented with a left supersript 0 (does not change the mass number) and a left underscript 1-  (the atomic number increases in 1).

4)  Then when iodine-131 emits a beta particle (electron) its mass number does not change, but the atomic number is increased from 53 to 54, becoming the isotope xenon - 131.

That is what the superscript to the left of Xe shows (131), while the subscript shows 54.

The subscript of electron is  - 1.

So the balance is:

Iodine              Xenon           electron

 131       =          131       +       0

   53       =            54       -         1

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