A) \[3.1\times {{10}^{22}}m{{s}^{-2}},\,1.3\times {{10}^{19}}m{{s}^{-2}}\]
B) \[3.3\times {{10}^{18}}m{{s}^{-2}},\,3.2\,\times {{10}^{16}}m{{s}^{-2}}\]
C) \[2.5\times {{10}^{22}}\,m{{s}^{-2}},\,1.4\times {{10}^{19}}m{{s}^{-2}}\]
D) \[2.5\times {{10}^{18}}m{{s}^{-2}},\,1.3\times {{10}^{16}}m{{s}^{-2}}\]
Correct Answer: C
Solution :
(c) \[2.5\text{ }\times \text{ }{{10}^{22}}\text{ }m{{s}^{-2}},1.4\times {{10}^{19}}\text{ }m{{s}^{-2}}\] Force of mutual attraction between an electron and a proton: \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{e}^{2}}}{{{r}^{2}}}\] \[=\frac{\left( 9\times {{10}^{9}} \right){{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{{{\left( {{10}^{-10}} \right)}^{2}}}=2.3\times {{10}^{-8}}N\] Acceleration of electron \[=\frac{F}{{{m}_{e}}}\] \[=\frac{2.3\times {{10}^{-8}}}{9\times {{10}^{-31}}}=2.5\times {{10}^{22}}m{{s}^{-2}}\] Acceleration of proton\[=\frac{F}{{{m}_{p}}}\] \[=\frac{2.3\times {{10}^{-8}}}{1.66\times {{10}^{-27}}}=1.4\times {{10}^{19}}m{{s}^{-2}}\]
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