The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 Å$ apart is,
$(m_{e} ≃ 9 \times 10^{- 31} kg, e = 1.6 \times 10^{- 19} C)$ (Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}$ )NEET 2020 (Phase 2)

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Recall the formula of electrostatic force interns of charge on electron and septation between them, $F = K \frac{e^{2}}{r^{2}}$ , now use Newton's second law, $m a = K \frac{e^{2}}{r^{2}}$

$\Rightarrow a = K \frac{e^{2}}{m r^{2}}$ ,

so, the acceleration, $a = 9 \times 10^{9} \frac{{(1.6 \times 10^{- 19})}^{2}}{{(1.6 \times 10^{- 10})}^{2} (9 \times 10^{- 31})}$

$\Rightarrow a = 10^{- 29} \times 10^{51} = 10^{22} m s^{- 2}$

Physics - Electrostatics Question with Solution | TestHub

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