Question

The fission process $\mathrm{n}+{}^{235}\mathrm{U}\to {}^{236}\mathrm{U}\to {}^{144}\mathrm{Ba}+{}^{89}\mathrm{Kr}+3\mathrm{n}$ converts $0.185\mathrm{u}$ of mass into the kinetic energy of the fission products. What is the total kinetic energy in $\mathrm{MeV}$ ?

Short answer

We find the kinetic energy to be equal to, $E_k=167\mathrm{MeV}$

Step-by-step solution

Given Information

The fission process $\mathrm{n}+{}^{235}\mathrm{U}\to {}^{236}\mathrm{U}\to {}^{144}\mathrm{Ba}+{}^{89}\mathrm{Kr}+3\mathrm{n}$converts $0.185\mathrm{u}$of mass into the kinetic energy of the fission products.

Calculation

Here the mass is getting converted to kinetic energy hence the kinetic energy will be equal to
$E_k=\Delta m{c}^2=0.185\times 1.6\times {10}^{-27}\times {\left(3\times {10}^8\right)}^2=2.66\times {10}^{-11}\mathrm{J}=1.67\times {10}^{-8}\mathrm{eV}=167\mathrm{MeV}$