MHT CET · Physics · Electrostatics
In hydrogen atom an electron revolves around a proton (in nucleus) at a distance ' \(r\) ' \(m\). the intensity of electric field due to the proton at distance ' \(r\) ' is \(5 \times 10^{11} \mathrm{NC}^{-1}\), the magnitude of force between the electron and proton is [charge on electron \(=1.6 \times 10^{-19} \mathrm{C}\) ]
- A \(4 \times 10^8 \mathrm{~N}\)
- B \(8 \times 10^8 \mathrm{~N}\)
- C \(4 \times 10^{-8} \mathrm{~N}\)
- D \(8 \times 10^{-8} \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(8 \times 10^{-8} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & F=q E=1.6 \times 10^{-19} \times 5 \times 10^{11} \\ & =8 \times 10^{-8} \mathrm{~N}\end{aligned}\)
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