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
Verified Correct Answer: B.
Physical Principles & Numerical Solution
The equation above is Beta-decay:
where \beta is an electron → \(^{0} _{-1}e\)
2\beta 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 [ \(^{222}_{88}Ra\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #43045)