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(D) The electrostatic force between an electron and a proton is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.Acco

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$10^{22} \; m/s^{2}$

Vedclass · Answer

(D) The electrostatic force between an electron and a proton is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.According to Newton's second law,the acceleration of the electron is $a_{e} = \frac{F}{m_{e}}$.Substituting the expression for force: $a_{e} = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{m_{e} r^{2}}$.Given $r = 1.6 \; \mathring{A} = 1.6 	imes 10^{-10} \; m$,$e = 1.6 	imes 10^{-19} \; C$,$m_{e} = 9 	imes 10^{-31} \; kg$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \; Nm^{2} C^{-2}$.$a_{e} = \frac{9 	imes 10^{9} 	imes (1.6 	imes 10^{-19})^{2}}{9 	imes 10^{-31} 	imes (1.6 	imes 10^{-10})^{2}}$.$a_{e} = \frac{9 	imes 10^{9} 	imes 2.56 	imes 10^{-38}}{9 	imes 10^{-31} 	imes 2.56 	imes 10^{-20}}$.$a_{e} = \frac{10^{9} 	imes 10^{-38}}{10^{-31} 	imes 10^{-20}} = \frac{10^{-29}}{10^{-51}} = 10^{22} \; m/s^{2}$.

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \; \mathring{A}$ apart is,$(m_{e} \simeq 9 \times 10^{-31} \; kg, e = 1.6 \times 10^{-19} \; C)$. (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$)

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