Question

An $\alpha$-particles is projected towards the following nucleus with same kinetic energy in 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)

Answer 2

The distance of closest approach ($r_0$) is given by $r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$. Since $K$ is constant, $r_0 \propto Z$. Thus, the maximum distance of closest approach occurs for the nucleus with the largest atomic number ($Z$). Comparing the given options, Au (Z=79) has the largest atomic number.