JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2026
- Medium
- MCQ
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 \(\left.=1.6 \times 10^{-19} C \right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \( 5.76\times10^{-14} \)
Step-by-step Solution
Detailed explanation
By mechanical energy conservation
\(( Me )_{ i }=( Me )_{ f }\)
\(P E_i+K E_i=P E_f+K E_f\)
\(0+7.9 \times 10^6 \times 1.6 \times 10^{-19}=\frac{ k (2 e )( Ze )}{ r }+0\)
\(r =\frac{9 \times 10^9 \times 2 \times\left(1.6 \times 10^{-19}\right)^2 \times 79}{7.9 \times 10^6 \times 1.6 \times 10^{-19}}=2.88 \times 10^{-14} m\)
For diameter ⇒ \(D=2 r=5.76 \times 10^{-14} m\)
\(( Me )_{ i }=( Me )_{ f }\)
\(P E_i+K E_i=P E_f+K E_f\)
\(0+7.9 \times 10^6 \times 1.6 \times 10^{-19}=\frac{ k (2 e )( Ze )}{ r }+0\)
\(r =\frac{9 \times 10^9 \times 2 \times\left(1.6 \times 10^{-19}\right)^2 \times 79}{7.9 \times 10^6 \times 1.6 \times 10^{-19}}=2.88 \times 10^{-14} m\)
For diameter ⇒ \(D=2 r=5.76 \times 10^{-14} m\)
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