A) a = 6, p = 4
B) a = 6, p = 0
C) a = 8, p = 6
D) a = 3, p = 3
Correct Answer: C
Solution :
Let number of \[\alpha \] particles decayed be x and number of \[\beta \] particles decayed be y. Then equation for the decay is given by \[_{92}{{U}^{235}}\xrightarrow[{}]{{}}x\alpha _{2}^{4}+y\beta _{-1}^{0}+Pb_{82}^{203}\] Equating the mass number on both sides \[235=4x+203\] ...(i) Equating atomic number on both sides \[92=2y-82\] ...(ii) Solving Eqs. (i) and (ii), we get \[x=8,y=6\] \[\therefore \]\[8\alpha \]particles and \[6\beta \]particles are emitted in disintegration.
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