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,$\left(m_{e} \simeq 9 \times 10^{-31} kg , e=1.6 \times 10^{-19} C \right)$
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )
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,$\left(m_{e} \simeq 9 \times 10^{-31} kg , e=1.6 \times 10^{-19} C \right)$
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )
- [NEET 2020]
- A
$10^{25} \;m / s ^{2}$
- B
$10^{24} \;m / s ^{2}$
- C
$10^{23} \;m / s ^{2}$
- D
$10^{22} \;m / s ^{2}$
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