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  • The distance of closest approach of an -particle fired towards a nucleus with momentum , is <img alt=

The distance of closest approach of an \alpha-particle fired towards a nucleus with momentum p, is r. If the momentum of the \alpha - particle is 2 p, then the corresponding distance of closest approach is:

Option: 1

r/2


Option: 2

2r


Option: 3

4r


Option: 4

r/4


Answers (1)

best_answer

At the distance of closed approach, r

\mathrm{K}=\frac{1}{4 \pi \varepsilon__{0}} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}} \Rightarrow \mathrm{r}=\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~K}}

where, Ze = charge of the nucleus

2e = charge of the alpha particle

K = kinetic energy of the alpha particle

\because \quad \mathrm{K}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}

Where, p is the momentum of \alpha - particle and m is the mass of the electron

\therefore \quad \mathrm{r}=\frac{2 \mathrm{Ze}^2 2 \mathrm{~m}}{4 \pi \varepsilon_0 \mathrm{p}^2}                or                \mathrm{r} \propto \frac{1}{\mathrm{p}^2}

\frac{\mathrm{r}^{\prime}}{\mathrm{r}}=\left(\frac{\mathrm{p}}{\mathrm{p}^{\prime}}\right)^2=\left(\frac{\mathrm{p}}{2 \mathrm{p}}\right)^2=\frac{1}{4} \Rightarrow \mathrm{r}^{\prime}=\frac{\mathrm{r}}{4}

Posted by

SANGALDEEP SINGH

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