COMEDK · Physics · 28. Atomic Physics
In the head-on collision of two alpha particles \(\alpha_1\) and \(\alpha_2\) with the gold nucleus, the closest approaches are 31.4 fermi and 94.2 fermi respectively. Then the ratio of the energy possessed by the alpha particles \(\alpha_2 / \alpha_1\) is:
- A \(3: 1\)
- B \(1: 9\)
- C \(1: 3\)
- D \(9: 1\)
Answer & Solution
Correct Answer
(C) \(1: 3\)
Step-by-step Solution
Detailed explanation
The distance of closest approach \(r_0\) for an alpha particle of kinetic energy \(K\) colliding head-on with a nucleus of atomic number \(Z\) is given by the formula \(r_0 = \dfrac{1}{4 \pi \epsilon_0} \dfrac{2Ze^2}{K}\). From this expression, it is evident that…
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