An α - particle with kinetic energy K is heading towards a fixed nucleus of atomic number Z. The distance of its closest approach is directly proportional to

Question

An α - particle with kinetic energy K is heading towards a fixed nucleus of atomic number Z. The distance of its closest approach is directly proportional to (A) (z/K) (B) √( barZ/ K) (C) (K/Z) (D) √(K/Z)

Tardigrade Admin · Accepted Answer

K=((. Z e .) (. Z e .)/4 π Ïµ0 r) r propto (Z/K)

Q. An $α -$ particle with kinetic energy $K$ is heading towards a fixed nucleus of atomic number $Z .$ The distance of its closest approach is directly proportional to

$\frac{z}{K}$

$\frac{Z ˉ}{K}$

$\frac{K}{Z}$

$\frac{K}{Z}$

$K = \frac{( Z e ) ( Z e )}{4 π ϵ _{0} r}$
$r \propto \frac{Z}{K}$

Q. An α− particle with kinetic energy K is heading towards a fixed nucleus of atomic number Z. The distance of its closest approach is directly proportional to

Kz​

KZˉ​​

ZK​

ZK​​

K=4πϵ0​r(Ze)(Ze)​ r∝KZ​

Q. An $α -$ particle with kinetic energy $K$ is heading towards a fixed nucleus of atomic number $Z .$ The distance of its closest approach is directly proportional to

$\frac{z}{K}$

$\frac{Z ˉ}{K}$

$\frac{K}{Z}$

$\frac{K}{Z}$

$K = \frac{( Z e ) ( Z e )}{4 π ϵ _{0} r}$
$r \propto \frac{Z}{K}$