JEE Mains · Physics · STD 12 - 13. Nuclei
7.9 MeV \( \alpha \)-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) __________m.
\(\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 Nm ^2 / C ^2\right.\) and electron charge \(\left.=1.6 \times 10^{-19} C \right]\)
- A \( 5.76\times10^{-14} \)
- B \( 1.44\times10^{-13} \)
- C \( 2.88\times10^{-14} \)
- D \( 1.69\times10^{-12} \)
Answer & Solution
Correct Answer
(A) \( 5.76\times10^{-14} \)
Step-by-step Solution
Detailed explanation
By mechanical energy conservation \(( Me )_{ i }=( Me )_{ f }\) \(P E_i+K E_i=P E_f+K E_f\) \(0+7.9 \times 10^6 \times 1.6 \times 10^{-19}=\frac{ k (2 e )( Ze )}{ r }+0\)…
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