The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6Å apart is, ( m_e=9 10^-31 kg ,e=1.6 10^-19C) (Take, , 14 _0=9 10^9 Nm^2C^-2) [NEET 202…

Question

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6Å apart is, ( m_e=9 10^-31 kg ,e=1.6 10^-19C) (Take, , 14 _0=9 10^9 Nm^2C^-2) [NEET 2020]

Accepted Answer

Correct answer: C. (c) Force of mutual attraction between the electron and proton. (when, r=1.6Å=1.6 10^-10 m) is given as F=9 10^9 e^2r^2=9 10^9 (1.6 10^-19)^2(1.6 10^-10)^2 =9 10^-9 N ∴ Acceleration of electron = Fm_e= 9 10^-99 10^-31=10^22 m/s^2

$The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6Å apart is, (m_{e} = 9 \times 10^{- 31} kg, e = 1.6 \times 10^{- 19} C) (Take, , \frac{1}{4 π ϵ _{0}} = 9 \times 10^{9} Nm^{2} C^{- 2}) [NEET 2020]$

Solution

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