Question 338: Atoms & Nuclei

Question

An $$\alpha $$-particle of energy $$5$$ $$MeV$$ is scattered through $${180^ \circ }$$ by a fixed uranium nucleus. The distance of closest approach is of the order of

Accepted Answer

KEY NOTE :
Distance of closest approach
$${r_0} = {{Ze\left( {2e} ight)} \over {4\pi {\varepsilon _0}E}}$$
Energy, $$E = 5 	imes {10^6} 	imes1.6 	imes {10^{ - 19}}J$$
$$	herefore$$ $${r_0} = {{9 	imes {{10}^9} 	imes \left( {92 	imes 1.6 	imes {{10}^{ - 19}}} ight)\left( {2 	imes 1.6 	imes {{10}^{ - 19}}} ight)} \over {5 	imes {{10}^6} 	imes 1.6 	imes {{10}^{ - 19}}}}$$
$$ \Rightarrow r = 5.2 	imes {10^{ - 14}}m = 5.3 	imes {10^{ - 12}}cm$$

Answer

$${10^{ - 12}}\,cm$$

Answer

$${10^{ - 10}}\,cm$$

Answer

$$1A$$

Answer

$${10^{ - 15a}}\,cm$$

Atoms & Nuclei

An $\alpha -particle of energy 5 MeV is scattered through {180^ \circ }$ by a fixed uranium nucleus. The distance of closest approach is of the order of

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