Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is
A. 14.4\(\frac { Z }{ V }\) Å
B. 14.4\(\frac { V }{ Z }\) Å
C. 1.44\(\frac { Z }{ V }\) Å
D. 14.4\(\frac { V }{ Z }\) Å
Answer:C
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is
A. 14.4\(\frac { Z }{ V }\) Å B. 14.4\(\frac { V }{ Z }\) Å C. 1.44\(\frac { Z }{ V }\) Å D. 14.4\(\frac { V }{ Z }\) Å