Verified Answer :

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

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