Sulphur -35 (34.96903 amu) emits a beta - pa | vedclass

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Question

(C) The nuclear reaction is: $^{35}_{16}S ightarrow ^{35}_{17}Cl + ^{0}_{-1}e + \bar{
u}$.Mass defect $(\Delta m) = 	ext{mass of } ^{35}S - 	ext{

Vedclass · Accepted Answer

$0.16758$

Vedclass · Answer

(C) The nuclear reaction is: $^{35}_{16}S ightarrow ^{35}_{17}Cl + ^{0}_{-1}e + \bar{
u}$.Mass defect $(\Delta m) = 	ext{mass of } ^{35}S - 	ext{mass of } ^{35}Cl$.$\Delta m = 34.96903 \ amu - 34.96885 \ amu = 0.00018 \ amu$.The energy released $(Q)$ is given by $Q = \Delta m 	imes 931 \ MeV/amu$.$Q = 0.00018 	imes 931 \ MeV = 0.16758 \ MeV$.

Sulphur $-35$ $(34.96903 \ amu)$ emits a $\beta -$ particle but no $\gamma -$ rays,the product is chlorine $-35$ $(34.96885 \ amu)$. The maximum energy emitted by the $\beta -$ particle is (in $MeV$)

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