COMEDK · Physics · 28. Atomic Physics
The distance of closest approach when an alpha particle of kinetic energy \(6.5 \text{ MeV}\) strikes a nucleus of atomic number 50 is
- A 0.221 pm
- B 0.0221 pm
- C 4.42 pm
- D 1.101 pm
Answer & Solution
Correct Answer
(B) 0.0221 pm
Step-by-step Solution
Detailed explanation
At closest approach, KE = PE: \(r_0 = \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{q_1 q_2}{K}\) \(q_1 = 2e\), \(q_2 = 50e\), \(K = 6.5\) MeV \(= 6.5 \times 10^6 \times 1.6 \times 10^{-19}\) J \(r_0 = \dfrac{(9 \times 10^9) \times 100 \times (1.6 \times 10^{-19})}{6.5 \times 10^6}\)…
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