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What is the shortest distance between nucleus of \mathrm{Al} and the point where an \alpha particle with kinetic energy Mev falls on the foil made up of aluminium? \mathrm{\left(k=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{c}^2\right)}

Option: 1

1.67 \times 10^{-14} \mathrm{~m}


Option: 2

4.68 \times 10^{-15} \mathrm{~m}


Option: 3

5.98 \times 10^{-13} \mathrm{\, m}


Option: 4

 none of these


Answers (1)

best_answer

\mathrm{K E =\frac{K Ze 2_e}{r}}

\mathrm{r =\frac{9 \times 10^9 \times 13 \times 2\left(1.6 \times 10^{-19}\right)^2}{8 \times 10^6 \times 1.6 \times 10^{-19}}}

\mathrm{ =46.8 \times 10^{-16}}

\mathrm{ =4.68 \times 10^{-15} \mathrm{m}}

Posted by

Pankaj Sanodiya

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