Atomic Structure - Result Question 24
24. With what velocity should an $\alpha$-particle travel towards the nucleus of a copper atom so as to arrive at a distance $10^{-13} m$ from the nucleus of the copper atom?
(1997 (C), 3M)
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Answer:
Correct Answer: 24. $(6.3 \times 10^6)$
Solution:
- When $\alpha$-particle stop at $10^{-13} m$ from nucleus, kinetic energy is zero, i.e. whole of its kinetic energy at the starting point is now converted into potential energy.
Potential energy of this $\alpha$-particle can be determined as
$ \begin{aligned} & PE=-\frac{Z _1 \times Z _2 e^{2}}{\left(4 \pi \varepsilon _0\right) r} \\ & \left(Z _1=+2, Z _2=+29, \varepsilon _0=8.85 \times 10^{-12} J^{-1} C^{2} m^{-1}\right. \text {, } \\ & \left.r=10^{-13} m\right) \\ & \Rightarrow \quad|PE|=\frac{2 \times 29 \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 10^{-13}} J \\ & = \quad 1.33 \times 10^{-13} J \\ & = \quad \text { kinetic energy of } \alpha \text {-particle at } t=0 \\ & \Rightarrow \quad KE=\frac{1}{2} m v^{2}=1.33 \times 10^{-13} \\ & \Rightarrow \quad v=\sqrt{\frac{2 \times 1.33 \times 10^{-13}}{4 \times 1.66 \times 10^{-27}}}=6.3 \times 10^{6} ms^{-1} \end{aligned} $
BETA
