JEE Mains · Physics · STD 12 - 12. atoms
If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is _________ m.
(Atomic number of gold \(=79\) and \(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9\)in SI units)
- A \(2.95 \times 10^{-14}\)
- B \( 2.95 \times 10^{-16} \)
- C \( 3.85 \times 10^{-16} \)
- D \( 3.85 \times 10^{-14} \)
Answer & Solution
Correct Answer
(A) \(2.95 \times 10^{-14}\)
Step-by-step Solution
Detailed explanation
Energy conservation \(K _{ i }+ U _{ i }= K _{ f }+ U _{ f }\) \(7.7 \times 10^6 \times 1.6 \times 10^{-19}+0\) \(=0+\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)\left(79 \times 1.6 \times 10^{-19}\right)}{ r }\) \(r =2.95 \times 10^{-14}\)
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