When slow neutrons are incident on a target containing 235 92 U, … - Vidyadip

Question

When slow neutrons are incident on a target containing $\frac{235}{92}\text{U},$ a possible fission reaction is $\frac{235}{92}\text{U}+_0^1\text{n}\rightarrow\frac{141}{56}\text{Ba}+\frac{92}{36}\text{Kr}+3_0\text{n}^1$
Estimate the amount of energy released using following data :
$\text{M}\Big(\frac{235}{92}\text{U}\Big)=235.04\text{amu}$
$\Big(\text{M}\frac{92}{36}\text{Kr}\Big)=91.926\text{amu}$
$\text{M}\Big(\frac{141}{56}\text{Ba}\Big)=140.91\text{amu}$
$\text{M}=1.0087\text{ amu}$

✓

Answer

$\text{Mass of }[_{92}\text{U}^{235}+_0\text{n}^1]={236.0487}\text{ amu}$
$\text{Mass of}[{_{56}\text{Ba}^{141}}+_{36}\text{Kr}^{92}+3_0\text{n}^1]={235.8621}\text{ amu}$
Difference in mass $={0.1866}\text{amu}$
Energy released $={(0.1866)}\times{931}={173.7}\text{MeV}$

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