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

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1025 m/s2

1024 m/s2

1023 m/s2

1022 m/s2

answer is D.

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Detailed Solution

F=e24πε0r2ae=Fmeae=e24πε0mer2ae=9×109×(1.6)2×10−389×10−31×(1.6)2×10−20ae=1022m/s2