Question

Consider the induced nuclear fission reaction:
${}^{235}_{92}\text{U} + n \rightarrow {}^{93}_{37}\text{Rb} + {}^{141}_{55}\text{Cs} + 2n$
where neutron momenta in both initial and final states are negligible. The ratio of the kinetic energies (KE) of the daughter nuclei,
$\frac{\text{KE}({}^{93}_{37}\text{Rb})}{\text{KE}({}^{141}_{55}\text{Cs})}$
is __________.

• A. $\frac{93}{141}$
• B. $\frac{141}{93}$
• C. 1
• D. 0

Answer 1

Answer: (B)

$\frac{141}{93}$

Answer 2

The correct option is (B) :$\frac{141}{93}$