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(C) The nuclear reaction for beta-minus decay is given by: $_{Z}X^{A} ightarrow _{Z+1}Y^{A} + _{-1}e^{0} + \bar{
u} + Q$.The $Q$-value of the react

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$(M_{x} - M_{y}) c^{2}$

Vedclass · Answer

(C) The nuclear reaction for beta-minus decay is given by: $_{Z}X^{A} ightarrow _{Z+1}Y^{A} + _{-1}e^{0} + \bar{
u} + Q$.The $Q$-value of the reaction is the energy released,which is given by the mass defect multiplied by $c^{2}$.The atomic mass $M_{x}$ of nucleus $X$ is related to its nuclear mass $m_{x}$ by $M_{x} = m_{x} + Zm_{e}$. Thus,$m_{x} = M_{x} - Zm_{e}$.The atomic mass $M_{y}$ of nucleus $Y$ is related to its nuclear mass $m_{y}$ by $M_{y} = m_{y} + (Z+1)m_{e}$. Thus,$m_{y} = M_{y} - (Z+1)m_{e}$.The energy released $Q$ is given by $Q = (m_{x} - m_{y} - m_{e})c^{2}$.Substituting the expressions for nuclear masses:$Q = [(M_{x} - Zm_{e}) - (M_{y} - (Z+1)m_{e}) - m_{e}]c^{2}$$Q = [M_{x} - Zm_{e} - M_{y} + Zm_{e} + m_{e} - m_{e}]c^{2}$$Q = (M_{x} - M_{y})c^{2}$.Therefore,the maximum energy of the emitted beta particle is $(M_{x} - M_{y})c^{2}$.

List-$I$	List-$II$
$(P)$ ${ }_{92}^{238} U \rightarrow{ }_{91}^{234} Pa$	$(1)$ $1 \alpha$ and $1 \beta^{+}$
$(Q)$ ${ }_{82}^{214} Pb \rightarrow{ }_{82}^{210} Pb$	$(2)$ $3 \beta^{-}$ and $1 \alpha$
$(R)$ ${ }_{81}^{210} Tl \rightarrow{ }_{82}^{206} Pb$	$(3)$ $2 \beta^{-}$ and $1 \alpha$
$(S)$ ${ }_{91}^{228} Pa \rightarrow{ }_{88}^{224} Ra$	$(4)$ $1 \alpha$ and $1 \beta^{-}$
	$(5)$ $1 \alpha$ and $2 \beta^{+}$

$A$ nucleus $X$ emits a beta particle to produce a nucleus $Y$. If their atomic masses are $M_{x}$ and $M_{y}$ respectively,the maximum energy of the beta particle emitted is (where $m_{e}$ is the mass of an electron and $c$ is the velocity of light):

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In a radioactive decay chain,the initial nucleus is ${}_{90}^{232}Th$. At the end,$6$ $\alpha$-particles and $4$ $\beta$-particles are emitted. If the final nucleus is ${}_{Z}^{A}X$,then $A$ and $Z$ are given by:

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