Question

dans la réaction ci - dessous, indiquez le nombre de neutrons libres formés
$_{98}^{252}cf\to_{44}^{108}ru + _{54}^{140}xe+?_{0}^{1}n$

Explanation:

Step1: Balance mass - number

The mass - number of Cf is 252. The sum of the mass - numbers of Ru and Xe and neutrons should equal 252. Mass - number of Ru is 108 and of Xe is 140. Let the number of neutrons be $x$. Then $108 + 140+x=252$.

Step2: Solve for $x$

$108 + 140+x=252$ simplifies to $248+x = 252$. Subtracting 248 from both sides gives $x=252 - 248=4$.

Answer:

4