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

Question

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 (A) 5.89 MeV (B) 6.89 MeV (C) 8.89 MeV (D) 7.89 MeV

Tardigrade Admin · Accepted Answer

(B . E .)y=(90 × 1 .008 + 138 × 1 .009 - 228 .03)u =1.932 u =1.932× 931.5 MeV Hence, Binding energy is per Nucleon is =(1 .932 × 931 .5/228) = 7.89 MeV

$5.89 M e V$

$6.89 M e V$

$8.89 M e V$

$7.89 M e V$

$(B . E .)_{y} = (90 \times 1.008 + 138 \times 1.009 - 228.03) u$
$= 1.932 u$
$= 1.932 \times 931.5 M e V$
Hence, Binding energy is per Nucleon is
$= \frac{1.932 \times 931.5}{228}$
$= 7.89 M e V$

5.89MeV

6.89MeV

8.89MeV

7.89MeV

(B.E.)y​=(90×1.008+138×1.009−228.03)u =1.932u =1.932×931.5MeV Hence, Binding energy is per Nucleon is =2281.932×931.5​ =7.89MeV

$5.89 M e V$

$6.89 M e V$

$8.89 M e V$

$7.89 M e V$

$(B . E .)_{y} = (90 \times 1.008 + 138 \times 1.009 - 228.03) u$
$= 1.932 u$
$= 1.932 \times 931.5 M e V$
Hence, Binding energy is per Nucleon is
$= \frac{1.932 \times 931.5}{228}$
$= 7.89 M e V$