NEET · Physics · STD 12 - 1. Electric charges and fields
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 \(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\) )
- A \(10^{25} \;m / s ^{2}\)
- B \(10^{24} \;m / s ^{2}\)
- C \(10^{23} \;m / s ^{2}\)
- D \(10^{22} \;m / s ^{2}\)
Answer & Solution
Correct Answer
(D) \(10^{22} \;m / s ^{2}\)
Step-by-step Solution
Detailed explanation
\(F=\frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}}\) \(a_{e}=\frac{F}{m_{e}}\)
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