Question

The closest distance of approach of an alpha particle travelling with a velocity ‘v’ towards $A{l}_{13}$ nucleus is ‘d’. The closest distance of approach of an alpha particle travelling with velocity ‘4v’ towards $F{e}_{26}$ nucleus is

• A. $\frac{d}{8}$
  
• B. $\frac{d}{4}$
  
• C. $\frac{d}{8}$
  
• D. $8d$

Answer 1

The correct option is C $\frac{d}{8}$


$\frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_0}\frac{Z{e}^2}{d}d\alpha \frac{Z}{v^2};\frac{d_2}{d_1}=\frac{Z_2}{Z_1}\times {\left[\frac{V_1}{V_2}\right]}^2=\frac{26\times V^2}{13\times 16V^2}{d}_2=\frac{d_1}{8}=\frac{d}{8}$