In a nuclear reaction, ²³⁵₉₂U absorbs a neutron and splits: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x , ¹₀n Find x and — JEE Physics
In a nuclear reaction, ${}^{235}_{92}U$ absorbs a neutron and splits:
$${}^{235}_{92}U + {}^1_0n \rightarrow {}^{141}_{56}Ba + {}^{92}_{36}Kr + x\,{}^1_0n$$
Find $x$ and the energy released if mass defect $\Delta m = 0.186$ u.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
▲ 18
Conservation of mass number: $235 + 1 = 141 + 92 + x \Rightarrow x = 3$
$$E = \Delta m \times 931.5 MeV/u = 0.186 \times 931.5 \approx 173 MeV$$
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Discussion (4)
SP
Is there a faster shortcut for this in the actual exam? Time is tight.
K
Underrated solution. The way you set it up makes it almost obvious.
L
Is there a faster shortcut for this in the actual exam? Time is tight.
J
Clean and to the point. Bookmarking this for revision.