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  4. The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 mathringA apart is, (me ≃ 9× 10-31 kg, e=1.6 × 10-19C) (Take (1/4πε0)=9× 109 Nm2 C-2)

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Q. 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 $\frac{1}{4\pi\varepsilon_{0}}=9\times 10^{9}\,Nm^{2}\,C^{-2})$

NEETNEET 2020Electric Charges and Fields

Solution:

$F=K \frac{e^{2}}{r^{2}}$
$a=K \frac{e^{2}}{mr^{2}}$
$a=9\times10^{9} \frac{\left(1.6\times10^{-19}\right)^{2}}{\left(1.6\times10^{-10}\right)^{2}\left(9\times10^{-31}\right)}$
$a=10^{-29}\times10^{51}=10^{22} m /s^{2}$