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GUJCET · Physics · Atoms
The distance of the closest approach of an alpha particle fired at a nucleus with Kinetic energy \(K\) is \(r_{0}\). The distance of the closest approach when the \(\alpha\) - particle is fired at the same nucleus with Kinetic energy 2 K will be _________ .
- A \(4 r_{0}\)
- B \(\frac{r_{0}}{2}\)
- C \(\frac{r_{0}}{4}\)
- D \(2 r_{0}\)
Answer & Solution
Correct Answer
(B) \(\frac{r_{0}}{2}\)
Step-by-step Solution
Detailed explanation
(B) \(\frac{r_{0}}{2}\)
Distance of closest approach
\(
\begin{aligned}
& r_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{Z} e^{2}}{\mathrm{~K}} \\
\therefore & r_{0} \propto \frac{1}{\mathrm{~K}} \\
& \frac{r_{01}}{r_{02}}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{2 \mathrm{~K}}{\mathrm{~K}} \\
\therefore & r_{02}=\frac{r_{01}}{2}=\frac{r_{0}}{2}
\end{aligned}
\)
Distance of closest approach
\(
\begin{aligned}
& r_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{Z} e^{2}}{\mathrm{~K}} \\
\therefore & r_{0} \propto \frac{1}{\mathrm{~K}} \\
& \frac{r_{01}}{r_{02}}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{2 \mathrm{~K}}{\mathrm{~K}} \\
\therefore & r_{02}=\frac{r_{01}}{2}=\frac{r_{0}}{2}
\end{aligned}
\)
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