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
Number of neutrons in a parent nucleus \[X\], which gives \[_{7}{{N}^{14}}\] nucleus after two successive \[\beta \] emissions would be [CBSE PMT 1998; MP PMT 2003]
A) 9
B) 8
C) 7
D) 6
Correct Answer:
A
Solution :
\[_{5}{{X}^{14}}{{\xrightarrow{-2\beta }}_{7}}{{N}^{14}}\] than no. of neutrons in \[_{5}{{X}^{14}}=14-5=9\].