A) \[{{10}^{-12}}cm\]
B) \[{{10}^{-16}}cm\]
C) \[{{10}^{-10}}cm\]
D) \[{{10}^{-14}}cm\]
Correct Answer: A
Solution :
: At the distance of closest approach d, Kinetic energy of a-particle = Potential energy of a-particle and gold nucleus i.e., \[K=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(2e)(Ze)}{d}=\frac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}d}\] or \[d=\frac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}K}\] Here, \[K=5MeV=5\times 1.6\times {{10}^{-13}}J\] \[(\because \,\,1MeV=1.6\times {{10}^{-13}})\] For gold, \[Z=79\] \[\therefore \] \[d=\frac{(2)(9\times {{10}^{9}}N\,{{m}^{2}}{{C}^{-2}})(79){{(1.6\times {{10}^{-19}}C)}^{2}}}{(5\times 1.6\times {{10}^{-13}}J)}\] \[=4.55\times {{10}^{-14}}m={{10}^{-12}}cm\]
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