Question

An $\alpha$ - particle is projected towards the following nucleus with same kinetic energy in different experiment the distance of closet approach is maximum for

• A. Na(Z=11)
• B. Ca(Z=20)
• C. Ag (Z=47)
• D. Au (Z=79)

Answer 1

Answer: (D)

Au (Z=79)

K.E $=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}$
$q_1$ = change on $1st$ particle
$q_2$ = charge on second particle
$r$ = distance between them or distance of closet approach
Assume $q_1$ charge on alpha particle
$={\alpha }_q$
$q_2$ = charge on nuclei given in question
$r=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{K.E}$
K.E is same for all nuclei. But $q_2$ is maximum for gold $(+79)$. Hence $r$ will be maximum for gold.