Question

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is :

(m_e = 9 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)

(Take $1\over4\pi\epsilon_o$ = 9 × 10⁹Nm²C⁻²)

(1) 10²⁵ m/s²

(2) 10²⁴ m/s²

(3) 10²³ m/s²

(4) 10²² m/s²

Answer 1

Answer (4): 10²² m/s²

Solution:

$F = {1\over 4\pi e_0} × {e^2\over r}$

= 9 × 10^-9

Acceleration of electron and proton = $F\over Me$

=10²² m/s²