Determine the value of x from the nuclear reaction below,

\(^{222}_{86}Rn\) → \(^{x}_{88}Ra\) + 2β + energy

A.

220

B.

222

C.

224

D.

226

E.

228

Correct answer is B

The equation above is Beta-decay:
where β is an electron → \(^{0} _{-1}e\)

2β in the equation = \(^{0} _{-1}e\) + \(^{0} _{-1}e\) →  \(^{0} _{-2}e\)

\(^{222}_{86}Rn\) → \(^{x}_{88}Ra\) + \(^{0} _{-2}e\) + energy

where  \(^{x}_{88}Ra\) + \(^{0} _{-2}e\) → \(^{222}_{86}Rn\)

: x = 222 for element Radon [ \(^{222}_{88}Ra\)