Question

Answer the attachment......




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Answer 1

$\large\rm { 1MeV = 1.602 × 10^{-13} Nm}$

At the closest distance of approach, K.E. will be 0.

Charge of $\large\rm {\bold{ \alpha}}$ - particle

$\large\rm { = 2 × 1.602 × 10^{-19} C}$

Charge of Uranium nucleus

$\large\rm { = 92 × 1.602 × 10^{-19} C}$

To calculate closest approach,

$\large\rm { \frac{1}{2} × 1.602 × 10^{-13} = }$

$\large\rm { \frac{9 × 10^{9} × (92×2) × (1.602 × 10^{-19})}{r}}$

$\large\rm { r = 10^{-12} cm}$

so, 10^-12 cm is your answer