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Q.

With what velocity should an $α$ -particle travel towards the nucleus of a copper atom so as to arrive at a distance of $10^{- 13} m$ from the nucleus of copper atom (At. No. of $Cu = 29$ ). If your answer in scientific notation is $x \times 10^{y}$ , then find the value of y.

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answer is 6.

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Detailed Solution

$\begin{array}{l} R = \frac{4 k Z e^{2}}{m_{α} \cdot v_{α}^{2}} \Rightarrow v_{α} = \sqrt{\frac{4 k Z}{m_{α} \cdot R}} \cdot e \\ = \sqrt{\frac{4 \times 9 \times 10^{9} \times 29}{4 \times 1.66 \times 10^{- 27} \times 10^{- 13}}} \times 1.6 \times 10^{- 19} \\ = 6.34 \times 10^{6} m / s \end{array}$

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With what velocity should an α-particle travel towards the nucleus of a copper atom so as to arrive at a distance of 10-13m from the nucleus of copper atom (At. No. of Cu=29 ). If your answer in scientific notation is x×10y, then find the value of y.