The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Ao apart is me=9×10−31kg, e=1.6×10−19C Take 14πεo=9×109 Nm2C−2

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

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$10^{3} m / s^{2}$

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

$10^{24} m / s^{2}$

$10^{25} m / s^{2}$

answer is D.

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Detailed Solution

Force of mutual attraction between the electron and proton is
$F = \frac{{Ke}^{2}}{r^{2}} (Here, K = 9 \times 109)$
$\Rightarrow F = 9 \times 10^{9} \times \frac{e^{2}}{r^{2}}$
$9 \times 10^{9} \times \frac{{(1.6 \times 10^{- 19})}^{2}}{{(1.6 \times 10^{- 10})}^{2}} = 9 \times 10^{- 9} N$
$∴$ Acceleration of electron $= \frac{F}{m_{e}} = \frac{9 \times 10^{- 9}}{9 \times 10^{- 31}}$
$= 10^{- 9} \times 10^{+ 31} = 10^{22} m / s^{2}$