Question

An $\alpha$-particle having kinetic energy of 8Me V is projected towards the nucleus of Z =22 .Calculate its distance of nearest approach.

Answer 1

$K.E=8MeV$

$P.E=\frac{kqq}{r}=\frac{k2e\times 22e}{r}$

$P.E=K.E$

$\frac{9\times {10}^9\times 44\times e^2}{r}=8\times {10}^6\times e$

$r=\frac{9\times {10}^9\times 44\times e}{8\times {10}^6}=7.930\times {10}^{-15}$

$r=0.0079pm$

$r=0.79pcm$

We get, $r=0.79pcm.$