A) 4, 2
B) 2, 2
C) 8, 6
D) 6, 4
Correct Answer: D
Solution :
\[_{90}T{{h}^{232}}\to {{\,}_{82}}P{{b}^{208}}+x{{\,}_{2}}H{{e}^{4}}+y{{\,}_{-1}}{{\beta }^{0}}\] Equating mass no. 232 = 208 + 4x + 0 y or 4x = 24 or x = 6 Equating atomic no. 90 = 82 + 2x ?y or 90 = 82 + 2 × 6 ? y or \[y=4\] Hence \[6\alpha \] and \[4\beta \] particles will be emitted.
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