CBSE 2022 · Region 4 · Set 1 · Q6 · 3 marks
(a)In Geiger-Marsden experiment, calculate the distance of closest approach for an alpha particle with energy $\displaystyle 2.56 \times 10^{-12} \mathrm{~J}$. Consider that the particle approaches gold nucleus $\displaystyle (\mathrm{Z}=79)$ in head-on position.(b)If the above experiment is repeated with a proton of the same energy, then what will be the value of the distance of closest approach ?
(a)
In Geiger-Marsden experiment, calculate the distance of closest approach for an alpha particle with energy $\displaystyle 2.56 \times 10^{-12} \mathrm{~J}$. Consider that the particle approaches gold nucleus $\displaystyle (\mathrm{Z}=79)$ in head-on position.
(b)
If the above experiment is repeated with a proton of the same energy, then what will be the value of the distance of closest approach ?
Marking-scheme solution
(a)
Calculation of distance of closest approach & $\displaystyle 1_{2}^{1}$
(b)
Calculation of distance of closest approach & $\displaystyle 1_{2}^{1}$
a) For $\displaystyle \alpha$ particles, distance of closest approach\begin{aligned}
r_{\alpha} & =\frac{1}{4 \pi \varepsilon_{0}} \frac{2 Z e^{2}}{E_{k}}
r_{\alpha} & =\frac{9 \times 10^{9} \times 2 \times 79 \times\left(1.6 \times 10^{-19}\right)^{2}}{2.56 \times 10^{-12}}
& =14.22 \times 10^{-15} \mathrm{~m}
& =14.22 \mathrm{fm}
\end{aligned}
b) For proton, distance of closest approach\begin{aligned}
r_{p} & =\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{E_{k}}
r_{p} & =\frac{r_{\alpha}}{2}
& =7 \cdot 11 \times 10^{-15} \mathrm{~m}
\end{aligned}
$$
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.