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  • The acceleration of an electron due to the mutual attraction between the electronand a proton when they are apart is, <img alt="\left

The acceleration of an electron due to the mutual attraction between the electronand a proton when they are 1.6 \AA apart is,

\left ( me\simeq 9 \times 10^{31} kg, e =1. 6 \times 10 ^{-19}C\right )    \left ( Take \frac{1}{4 \pi \epsilon _0}=9 \times 10^9 N m^2 C^{-2}\right )

 

Option: 1

10^{23} m/s^2


Option: 2

10^{22} m/s^2


Option: 3

10^{25} m/s^2

 


Option: 4

10^{24} m/s^2


Answers (1)

best_answer

F= \frac{1}{4 \pi \varepsilon _0}\times \frac{q_1q_2}{r^2}

F= \frac{9\times 10^9 \times (1.6 \times 10^{-19})^2}{(1.6 \times 10^{-19})^2}

F=9 \times 10^{-9}N

Acceleration of electron

a_e=\frac{F}{m_e}=\frac{9\times 10^{-9}}{9 \times 10^{31}}kg

a_e=10^{22}m/s^2

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manish

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