Write the fission reaction condition

A neutron strikes uranium-235: ${}_0^1n + {}_{92}^{235}U \rightarrow \text{fragments}$

So, total mass number on the left is: $1+235=236$ and total atomic number on the left is: $0+92=92$

For any correct fission fragment set, both must be conserved:

• Mass number $A = 236$
• Atomic number $Z = 92$

Check option A

${}_{56}^{140}Xe + {}_{38}^{94}Sr + 3{}_0^1n$

• Mass number: $140+94+3=237$
• Atomic number: $56+38=94$

This does not satisfy conservation.

Hence, A is incorrect.

Check option B

${}_{51}^{153}Sb + {}_{41}^{99}Nb + 3{}_0^1n$

• Mass number: $153+99+3=255$
• Atomic number: $51+41=92$

Atomic number is correct, but mass number is not.

Hence, B is incorrect.

Check option C

${}_{56}^{144}Ba + {}_{36}^{89}Kr + 4{}_0^1n$

• Mass number: $144+89+4=237$
• Atomic number: $56+36=92$

Atomic number is correct, but mass number is not.

Hence, C is incorrect.

Check option D

${}_{56}^{144}Ba + {}_{36}^{89}Kr + 3{}_0^1n$

• Mass number: $144+89+3=236$
• Atomic number: $56+36=92$

Both mass number and atomic number are conserved.

Hence, D is correct.

Final Answer

The correct nuclear fission fragments are: ${}_{56}^{144}Ba + {}_{36}^{89}Kr + 3{}_0^1n$

So the correct option is D.