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  4. A nuclide X , initially at rest, undergoes α -decay according to the equation 92ZA arrow ZY228+α .The α -particle produced in the above process is found to move in a circular track of radius 0.11 m in a uniform magnetic field of 3T . Find the binding energy per nucleon (in MeV ) of the daughter nuclide Y . Given that m(Y)=228.03 u;m(2He4)=4.003 u m(0n1)=1.009u;m(H1)=1.008 u u=1 amu=931.5 MeV c- 2=1.66× 10- 27kg

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Q. A nuclide $X$ , initially at rest, undergoes $\alpha $ -decay according to the equation ${}_{92}^{}Z_{}^{A} \rightarrow {}_{Z}^{}Y_{}^{228}+\alpha $ .The $\alpha $ -particle produced in the above process is found to move in a circular track of radius $0.11\,m$ in a uniform magnetic field of $3T$ . Find the binding energy per nucleon (in $MeV$ ) of the daughter nuclide $Y$ . Given that
$m\left(Y\right)=228.03\,u;m\left({}_{2}^{}He_{}^{4}\right)=4.003\,u$
$m\left({}_{0}^{}n_{}^{1}\right)=1.009u;m\left(H^{1}\right)=1.008\,u$
$u=1\,amu=931.5\,MeV \,c^{- 2}=1.66\times 10^{- 27}kg$

NTA AbhyasNTA Abhyas 2022

Solution:

$\left(B . E .\right)_{y}=\left(90 \times 1 .008 + 138 \times 1 .009 - 228 .03\right)u$
$=1.932\,u$
$=1.932\times 931.5\,MeV$
Hence, Binding energy is per Nucleon is
$=\frac{1 .932 \times 931 .5}{228}$
$= \, 7.89\,MeV$