Answer to Question #180249 in Atomic and Nuclear Physics for ary

Question

One example of a nuclear fission reaction involving slowly moving neutrons is

1/0n + 235/92 U ---> 236/92 U* ---> 95/42 Mo + 139/57 La + (2) 1/0n

What is the total kinetic energy (in eV) of the products of the reaction? The relevant masses are:

235/92 U = 235.043 924 u; 95/42 Mo - 94.9058 u; 139/59 La = 138.9061 u, and 1/0n = 1.0087u

Answer 1

To be given in question

Urenium mass =235.043$u$

Molebidnum mass =94.9058$u$

Neutron mass =1.0087$u$

Lanthanum mass =13 8.9061 $u$

To be asked in question

Kinetic energy in eV(election volt) =?

²³⁵₉₂U +¹n $\longrightarrow$ ²³⁶U^*$\longrightarrow$ ⁹⁵₄₂ Mo+¹³⁹₅₇La+2(¹n)

$M_{1}=M({235_{U}})+M_{{1}_{n}}$

$M_{2}=M_{{235}_{U}}+M_{{139}_{La}}+2M_{{1}_{n}}$

$\Delta M =M_{1}-M_{2}$

$M_{1}= 235.04+1.0087$

$M_{1}=236.0487 u$

$M_{2}=(94.9058+138.9061+2\times1.0087) u$ $M_{2}=235.8293 u$

$\Delta M =M_{1}-M_{2}$

Put value

$\Delta M$ =236.0487-235.8293 $u$

$\Delta M=0.2194 u$

$\Delta E=\Delta Mc^2$

$\Delta E=0.2194\times931.5$

$\Delta E=204.3711 mev$

1kg ²³⁵U fission energy release

$K.E=\frac{6.02\times 10^{23}}{0.235}\times{204.3711}$

$K.E=5.2353\times 10^{26}Mev$$K.E=5.2353\times 10^{23}Kev$