Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन
JEE Main & Advanced
Chemistry
Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन
Question Bank
Causes of radioactivity and Group displacement law
question_answer
\[_{89}A{{c}^{231}}\] gives \[_{82}P{{b}^{207}}\] after emission of some \[\alpha \]and \[\beta \]-particles. The number of such \[\alpha \] and \[\beta \]-particles are respectively [MP PMT 1993; UPSEAT 2001]
A) 5, 6
B) 6, 5
C) 7, 5
D) 5, 7
Correct Answer:
B
Solution :
Number of a-particles = \[\frac{231-207}{4}=6\] Number of b-particles \[=89-82-2\times 6=5\].