Practice portal › Nuclei › Nuclear Reactions, Fission and Fusion

Find the value of x and y from below given nuclear reaction ²³⁵₉₂U+¹₀n→¹³³ₓSb+^y₄₁Nb+4 ¹₀n

Asked in GUJCET 2024 · Q-value, decay and conservation laws

Answer: (2) (51, 99)

Step-by-step solution

Idea: charge and nucleon number are both conserved, which gives one equation each.

Charge: 92+0=x+41+0, so x=51 — the fragment is ¹³³₅₁Sb, antimony.

Nucleons: 235+1=133+y+4, so y=236-137=99 — the other fragment is ⁹⁹₄₁Nb.

Hence (x,y)=(51, 99).

Why the other options are wrong

More Nuclear Reactions, Fission and Fusion questionsAll Nuclear Reactions, Fission and Fusion questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer