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13.Nuclei
medium
An alpha particle is projected towards a stationary ${}_{92}^{235}U$ nucleus with $KE$ kinetic energy find distance of closest approach
A
$\frac{{K{e^2}}}{{\left( {KE} \right)}}$
B
$\frac{{92\,K{e^2}}}{{\left( {KE} \right)}}$
C
$\frac{{K{e^2}}}{{92\,\left( {KE} \right)}}$
D
$\frac{{KE}}{{K{e^2}}}$
Solution
Energy conservation
Total Energy initial $=$ Total energy final
Hence $\mathrm{KE}=\frac{\mathrm{K}(\mathrm{ze}) \mathrm{e}}{\mathrm{d}}=\mathrm{PE}$
as $z = 92$ $\boxed{d = 92\frac{{{\text{K}}{{\text{e}}^2}}}{{({\text{KE}})}}}$
Standard 12
Physics
