# After absorbing a neutron, a Uranium-235 nucleus can fission to produce Barium-141 and krypton-92. Calculate the energy that would be released from this fission of a Uranium-235 atom. How do you work this out?

May 8, 2017

$1.81 \cdot {10}^{17} J$

#### Explanation:

To solve this question we require some extra information which is not provided but I will take variables in place of them .

${1}^{s t}$ method :

Mass defect method
we must be given the mass of protons and that of neutrons which is not provided so we will take the values provided over internet .

$1 \text{Proton"=1.00727"amu}$
$1 \text{Neutron"=1.00866 "amu}$

Net mass of ${U}^{235} = 92 \cdot 1.00727 + 143 \cdot 1.00866 = 236.90722 U$

similarly we calculate the mass of barium and krypton ,
$B {a}^{141} = 56 \cdot 1.00727 + 85 \cdot 1.00866 = 142.14322 U$
$K {r}^{92} = 36 \cdot 1.00727 + 56 \cdot 1.00866 = 92.74668 U$

Mass defect($\Delta M$) =$236.90722 - 142.14322 - 92.74668 = 2.01732 U$
now according to Einsteins mass energy relation we know

$E = M {c}^{2}$

it can be written as $E = \Delta M {c}^{2} = \left(M \left({U}^{235}\right) - M \left(B {a}^{141}\right) - M \left(K {r}^{92}\right)\right) {c}^{2} = 2.01732 \cdot \left(3 \cdot {10}^{8}\right) = 1.81 \cdot {10}^{17} J$
this is the case when both the daughter nuclei are not radioactive but if they are we subtract the mass of two neutrons and then multiply by ${c}^{2}$.

May 8, 2017

For the reaction stated, the mass defect is:

$\Delta m = \left({m}_{U} + {m}_{n}\right) - \left({m}_{B a} + {m}_{K r}\right)$

Using rounded amu figures found on Wikipedia:

$= \left(\left(235.044 + 1.008\right) - \left(140.914 + 91.926\right)\right)$

$\approx 3.212 \setminus \text{amu}$

A mass of 1 amu converts to energy via $E = m {c}^{2}$ as: $1 \text{amu} \approx 931 M e V$

So here:

$E \approx 3 G e V$

That's about $4.8 \times {10}^{- 7} J$