JEE MAIN - Physics (2004 - No. 7)
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
$${10^{ - 12}}\,cm$$
$${10^{ - 10}}\,cm$$
$$1A$$
$${10^{ - 15a}}\,cm$$
Explanation
KEY NOTE :
Distance of closest approach
$${r_0} = {{Ze\left( {2e} \right)} \over {4\pi {\varepsilon _0}E}}$$
Energy, $$E = 5 \times {10^6} \times1.6 \times {10^{ - 19}}J$$
$$\therefore$$ $${r_0} = {{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)} \over {5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}$$
$$ \Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm$$
Distance of closest approach
$${r_0} = {{Ze\left( {2e} \right)} \over {4\pi {\varepsilon _0}E}}$$
Energy, $$E = 5 \times {10^6} \times1.6 \times {10^{ - 19}}J$$
$$\therefore$$ $${r_0} = {{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)} \over {5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}$$
$$ \Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm$$
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