Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

When a uranium isotope $_{92}^{235}\mathrm{U}$ is bombarded with a neutron, it generates $_{36}^{89}\mathrm{Kr}$ , three neutrons and:

$_{40}^{91}\mathrm{Zr}$

$_{36}^{101}\mathrm{Kr}$

$_{36}^{103}\mathrm{Kr}$

$_{56}^{144}\mathrm{Ba}$

Conservation Laws: In any nuclear reaction, both the mass number ( $A$ ) and the atomic number ( $Z$ ) must be conserved.
Reaction Equation: The given fission reaction can be written as: $_{92}^{235}\mathrm{U} + _{0}^{1}\mathrm{n} \rightarrow _{36}^{89}\mathrm{Kr} + 3_{0}^{1}\mathrm{n} + _{Z}^{A}\mathrm{X}$
Balancing Mass Number ( $A$ ): $235 + 1 = 89 + 3(1) + A$ $236 = 92 + A \implies A = 236 - 92 = 144$
Balancing Atomic Number ( $Z$ ): $92 + 0 = 36 + 3(0) + Z$ $92 = 36 + Z \implies Z = 92 - 36 = 56$
Identification: The element with atomic number $Z=56$ is Barium (Ba). Thus, the missing nucleus is $_{56}^{144}\mathrm{Ba}$ .

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