Question

the distance of closest approach of an alpha -particles fired towards a nucleus with momentum p is r. what will be the distance of closest approach when the momentum of the alpha -particles is 2p????????solve this question

Answer 1

this is your answer.......

Answer 2

At the distance of closest approach, all of the kinetic energy of incident particle is converted to potential energy of the system.

$Therefore,\quad { E }_{ i }\quad =\quad \frac { { P }^{ 2 } }{ 2m }$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 4\pi { \varepsilon  }_{ o } } \quad \frac { Ze\quad \times \quad q }{ r }$

$Here,\\ \quad \quad \quad \quad r\quad =\quad distance\quad of\quad closed\quad approach$

$\quad \quad \quad \quad q\quad =\quad charge\quad of\quad incident\quad particle$

$Therefore,\\ \quad for\quad particles\quad with\quad \quad different\quad momenta,\quad r\quad \alpha \quad \frac { 1 }{ { P }^{ 2 } }$

$Hence,\quad for\quad '\alpha '\quad particle\quad with\quad momentum\quad 3P,$$The\quad distance\quad of\quad closest\quad approach\quad becomes\quad \frac { d }{ { 3 }^{ 2 } } \quad =\quad \frac { d }{ 9 }$