The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 A0 apart is, take, 14πε0=9×109Nm2C-2,me≃9×10-31 kg,e=1.6×10-19

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 $\overset{0}{A}$ apart is, $(\begin{array}{l} take, \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2}, m_{e} ≃ 9 \times 10^{- 31} kg, \\ e = 1.6 \times 10^{- 19} \end{array})$

A
$10^{24} m / s^{2}$
B
$10^{23} m / s^{2}$
C
$10^{22} m / s^{2}$
D
$10^{25} m / s^{2}$

Solution:

Force due to mutual attraction between the electron and proton. (when, $r = 1.6 \overset{0}{A} = 1.6 \times 10^{- 10} m$ ) is given as

$F = 9 \times 10^{9} \times \frac{e^{2}}{r^{2}}$

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

$∴$ Acceleration of electron

$= \frac{F}{m_{e}} = \frac{9 \times 10^{- 9}}{9 \times 10^{- 31}} = 10^{22} m / s^{2}$

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