In the given nuclear reaction 21584Po →21182Pb + X \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaa % baaaaaaaaapeGaaGioaiaaisdaa8aabaWdbiaaikdacaaIXaGaaGyn % aaaakiaadcfacaWGVbGaeyOKH46damaaDeaaleaapeGaaGioaiaaik % daa8aabaWdbiaaikdacaaIXaGaaGymaaaakiaadcfacaWGIbGaey4k % aSIaamiwaaaa!4578! {}_{84}^{215}Po \to {}_{82}^{211}Pb + X\), what is X?

In the given nuclear reaction 21584Po →21182Pb + X \(% MathType!MTEF

1.	In the given nuclear reaction 21584Po →21182Pb + X \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaa % baaaaaaaaapeGaaGioaiaaisdaa8aabaWdbiaaikdacaaIXaGaaGyn % aaaakiaadcfacaWGVbGaeyOKH46damaaDeaaleaapeGaaGioaiaaik % daa8aabaWdbiaaikdacaaIXaGaaGymaaaakiaadcfacaWGIbGaey4k % aSIaamiwaaaa!4578! {}_{84}^{215}Po \to {}_{82}^{211}Pb + X\), what is X?
A.	Neutron
B.	γ - Particle
C.	α - Particle
D.	β - Particle
Answer» D. β - Particle