In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number 80 , then maximum velocity of $\alpha$-particle is __________ $\times 10^5 \mathrm{~m} / \mathrm{s}$ approximately.
($\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}$ unit, mass of $\alpha$ particle $=6.72 \times 10^{-27} \mathrm{~kg}$)
Answer (integer)
156
Solution
<p>$$\begin{aligned}
& \frac{1}{2} m v_0^2=\frac{1}{4 \pi \epsilon_0} \frac{z(e)}{r} \\
& \frac{1}{2} \times 6.72 \times 10^{-27} v_0^2=9 \times 10^9 \times \frac{80 \times 2 \times 1.6 \times 1.6 \times 10^{-19} \times 10^{-19}}{4.5 \times 10^{-14}} \\
& v_0^2=\frac{2 \times 9 \times 80 \times 1.6 \times 1.6 \times 2}{6.72 \times 4.5} \times 10^{-38+14+27+9} \\
& v_0^2=243.8 \times 10^{12} \\
& v_0=15.6 \times 10^6 \\
& v_0=156 \times 10^5
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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