In Geiger-Marsden experiment, if the initial speed of \(\alpha\) particle is doubled, then the closest distance of approach of the \(\alpha\) particle from the gold nucleus
- (Assume \(\alpha\) particle is projected straight towards the gold nucleus)
Show Hint
- becomes doubled
- becomes quadrupled
- becomes half
- becomes one-fourth
The Correct Option is D
Solution and Explanation
Step 1: Closest distance relation
\[ r_{\min} \propto \frac{1}{K} \]
Step 2: Relation between kinetic energy and speed
\[ K = \frac{1}{2}mv^2 \]
If speed is doubled:
\[ K' = \frac{1}{2}m(2v)^2 = 4K \]
Step 3: Effect on closest distance
\[ r'_{\min} \propto \frac{1}{K'} = \frac{1}{4K} \]
\[ r'_{\min} = \frac{r_{\min}}{4} \]
Final Answer: Becomes one-fourth
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