Atomic Structure 1 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 $10^{-13} \mathrm{~m}$ from the nucleus of the copper atom?
(1997 (C), 3M)
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Solution:
- When $\alpha$-particle stop at $10^{-13} \mathrm{~m}$ 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
$$ \begin{aligned} & \mathrm{PE}=-\frac{Z_{1} \times Z_{2} e^{2}}{\left(4 \pi \varepsilon_{0}\right) r} \ & \left(Z_{1}=+2, Z_{2}=+29, \varepsilon_{0}=8.85 \times 10^{-12} \mathrm{~J}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-1}\right. \text {, } \ & \left.r=10^{-13} \mathrm{~m}\right) \ & \Rightarrow \quad|\mathrm{PE}|=\frac{2 \times 29 \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 10^{-13}} \mathrm{~J} \ & =1.33 \times 10^{-13} \mathrm{~J} \ & =\text { kinetic energy of } \alpha \text {-particle at } t=0 \ & \Rightarrow \quad \mathrm{KE}=\frac{1}{2} m v^{2}=1.33 \times 10^{-13} \ & \Rightarrow \quad v=\sqrt{\frac{2 \times 1.33 \times 10^{-13}}{4 \times 1.66 \times 10^{-27}}}=6.3 \times 10^{6} \mathrm{~ms}^{-1} \end{aligned} $$
Topic 2 Advanced Concept (Quantum Mechanical Theory) Electronic Configuration and Quantum Number
