In a radioactive decay chain reaction, { }_{9 | vedclass

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Question

$Th _{90}^{230} ightarrow Po _{24}^{214}+ n _2^4+ m \beta_{-1}^0$

$230=214+4 n$

$n =\frac{16}{4}=4$

$90=84+ n 	imes 2- m 	imes 1$

$90=

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$Th _{90}^{230} ightarrow Po _{24}^{214}+ n _2^4+ m \beta_{-1}^0$

$230=214+4 n$

$n =\frac{16}{4}=4$

$90=84+ n 	imes 2- m 	imes 1$

$90=84+4 	imes 2- m 	imes 1$

$m =92-90=2$

Hence$\frac{ n }{ m }=\frac{4}{2}=2$ Ans.

Hence $\frac{ n }{ m }=\frac{4}{2}=2$ Ans.

column $I$	column $II$
$(A.)$Nuclear fusion	$(P.)$ Absorption of thermal neutrons by ${ }_{92}^{213} U$
$(B.)$Fission in a nuclear reactor	$(Q.)$ ${ }_{27}^{60} Co$ nucleus
$(C.)$ $\beta$-decay	$(R.)$ Energy production in stars via hydrogen conversion to helium
$(D.)$ $\gamma$-ray emission	$(S.)$ Heavy water
	$(T.)$ Neutrino emission

In a radioactive decay chain reaction, ${ }_{90}^{230} Th$ nucleus decays into ${ }_{84}^{214} Po$ nucleus. The ratio of the number of $\alpha$ to number of $\beta^{-}$particles emitted in this process is. . . . .

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