A) 0.59 nm
B) \[0.59\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) 5.9pm
D) 0.59pm
Correct Answer: D
Solution :
Distance of closest approach \[{{r}_{0}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{2\varepsilon {{e}^{2}}}{KE}\] \[\therefore \] \[{{r}_{0}}=\frac{9\times {{10}^{9}}\times 2\times 82\times {{(1.6\times {{10}^{-19}})}^{2}}}{400\times {{10}^{3}}\times 1.6\times {{10}^{-19}}}\] \[=\frac{9\times 164\times 1.6\times {{10}^{-15}}}{4}\] \[{{r}_{0}}=590\times {{10}^{-15}}m\] \[=0.59\times {{10}^{-12}}m\] \[=0.59pm.\]
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