A) \[\alpha =6,~\beta =4\]
B) \[\alpha =6,~\beta =0\]
C) \[\alpha =8,~\beta =6\]
D) \[\alpha =3,~\beta =3\]
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 \[92=2x-y+82\] ... (ii) Solving Eqs. (i) and (ii), we get \[x=8,y=6\] \[\therefore \] \[8\,\alpha \] particles and 6p particles are emitted in disintegration.
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