Atoms & Nuclei

JEE Main level, single correct, medium

160 of 239

Question 160 of 239

7.9MeVα7.9 \mathrm{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. [14πϵo=9×109Nm2/C2\left[\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right. and electron charge =1.6×10−19C]\left.=1.6 \times 10^{-19} \mathrm{C}\right]

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Answer: (B) 5.76×10−145.76 \times 10^{-14}

For Rutherford scattering, at the closest approach all kinetic energy converts to electrostatic potential energy.
For alpha particle-nucleus collision:
K.E.=14πε0Z1Z2e2rminK.E. = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r_{min}}
Where: Z1=2Z_1 = 2 (alpha), Z2=79Z_2 = 79 (gold), K.E.=7.9K.E. = 7.9 MeV
rmin=9×109×2×79×(1.6×10−19)27.9×106×1.6×10−19r_{min} = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.9 \times 10^6 \times 1.6 \times 10^{-19}}
rmin=2.88×10−14r_{min} = 2.88 \times 10^{-14} m
The estimated nuclear diameter is approximately:
d=2×rmin=5.76×10−14d = 2 \times r_{min} = 5.76 \times 10^{-14} m

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