Question

Which isotope will spontaneously decay and emit particles with a charge of +2?

(1) 53Fe (3) 198Au
(2) 137Cs (4) 220Fr

Answer 1

It is 220Fr that would spontaneously decay and emit particles with a charge of +2.

Answer 2

Answer: The correct option is 4.

Explanation: All the options will undergo some type of radioactive decay processes. There are 3 decay processes:

1) Alpha decay: It is a decay process in which alpha particle is released which has has a mass number of 4 and a charge of +2.

`_Z^A\textrm{X}\rightarrow _(Z-2)^(A-4)\textrm{Y}+_2^4\alpha`

2) Beta-minus decay: It is a decay in which a beta particle is released. The beta particle released has a mass number of 0 and a charge of (-1).

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

3) Beta-plus decay: It is a decay process in which a positron is released. The positron released has a mass number of 0 and has a charge of +1.

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

For the given options:

Option 1: This nuclei will undergo beta-plus decay process to form
`_(25)^(53)\textrm{Mn}`

`_(26)^(53)\textrm{Fe}\rightarrow _(25)^(53)\textrm{Mn}+_(+1)^0\beta`

Option 2: This nuclei will undergo beta-minus decay process to form
`_(80)^(198)\textrm{Hg}`

`_(79)^(198)\textrm{Au}\rightarrow _(80)^(198)\textrm{Hg}+_(-1)^0\beta`

Option 3: This nuclei will undergo a beta minus decay process to form
`_(56)^(137)\textrm{Ba}`

`_(55)^(137)\textrm{Cs}\rightarrow _(56)^(137)\textrm{Ba}+_(-1)^0\beta`

Option 4: This nuclei will undergo an alpha decay process to form
`_(85)^(216)\textrm{At}`

`_(87)^(220)\textrm{Fr}\rightarrow _(85)^(216)\textrm{At}+_2^4\alpha`

Hence, the correct option is 4.