E.4 Fission

This question describes induced fission — a type of fission triggered by neutron absorption. The neutron makes the nucleus unstable, initiating a specific sequence of events.

When a fissile nucleus like plutonium-239 absorbs a slow neutron, it briefly forms an unstable intermediate nucleus (plutonium-240 in this case). This intermediate is so energetically unstable that it almost immediately splits into two fission fragments plus additional neutrons.

The correct sequence is: neutron absorbed → unstable intermediate nucleus formed → nucleus splits into two smaller nuclei (fission fragments) + neutrons. This matches the option stating the nucleus becomes a highly unstable intermediate that rapidly splits.

The question asks what happens immediately after a fissile nucleus absorbs a slow neutron in induced fission.

"The nucleus emits an alpha particle and becomes stable" describes alpha decay, a completely different nuclear process. Induced fission does not proceed via alpha emission.

"The nucleus undergoes beta decay before splitting" is incorrect. Beta decay is a separate decay mode; induced fission does not require beta decay as a precursor step.

"The nucleus emits gamma radiation and returns to its ground state without splitting" describes nuclear de-excitation, not fission. While gamma radiation is also produced in fission, the nucleus does not simply stabilise — it splits.

The remaining option — that the nucleus becomes a highly unstable intermediate that rapidly splits — correctly describes the induced fission mechanism and is confirmed by the notes' description of U-235 becoming U-236 before splitting.

In any nuclear equation, both nucleon number (A) and proton number (Z) must be conserved. We need to find the A and Z of nucleus X.

Left side: $1+235=236$. Right side: $141+A_X+3(1)=144+A_X$. Setting equal: $A_X=236-144=92$.

Left side: $0+92=92$. Right side: $56+Z_X+3(0)=56+Z_X$. Setting equal: $Z_X=92-56=36$.

Nucleus X has $A=92$ and $Z=36$. The element with proton number 36 is krypton (Kr). Therefore X is ${}_{36}^{92}\text{Kr}$.

We must identify the missing fission product using conservation of both nucleon number and proton number.

From nucleon conservation: $A_X=236-141-3=92$. From proton conservation: $Z_X=92-56=36$. So X must have $A=92$, $Z=36$.

"${}_{36}^{91}Kr$" has $A=91$ (not 92) — eliminated. "${}_{36}^{93}Kr$" has $A=93$ (not 92) — eliminated.

"${}_{38}^{92}Sr$" has $Z=38$ (strontium, not krypton) — this does not satisfy proton conservation — eliminated.

Only "${}_{36}^{92}Kr$" satisfies both $A=92$ and $Z=36$, confirming it as the correct answer.