Determine the value of x from the nuclear reaction below,
\(^{222}_{86}Rn\) → \(^{x}_{88}Ra\) + 2β + energy
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 [ \(^{222}_{88}Ra\)
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