Question

alpha-particles are projected towards the following metals with the same kinetic energy. Towards which metal,the distance of closest approach is minimum ?
A)Cu (Z=29). B)Ag (Z=47)
C)Au (Z=79). D)CA (Z=20)

Answer 1

Answer:

A) cu (z=29)

Explanation:

thus value corrseponds the cu

Answer 2

Given:

$\alpha$ particles are radiated toward metal with the same KE.

To Find:

Distance for approach is minimum

$r_{min} = ?$

Solution:

Distance for Closest Approach

KE = PE

$KE = \dfrac{ kq_{1} q_{2}}{r}$

$r = \dfrac{kq_{1}q_{2} }{KE}$

$q_{1}$ = Charge of nuclei

$q_{2}$ = Charge on $\alpha$ particle

$r$ is directly proportional to charge

As Charge increase r increase vice versa r decrease on decreasing charge

So in this question Copper with atomic no. 29 has lowest charge so its distance of closest approach is minimum

Option A is the answer of this question