Question

If a proton and an electron are released when they are 2.00×10−10 m apart (typical atomic distances), find the initial acceleration of each of them.

Answer 1

q = magnitude of charge on each electron and proton = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

M = mass of proton = 1.67 x 10⁻²⁷ kg

r = distance between electron and proton = 2 x 10⁻¹⁰ m

F = electric force of attraction between electron and proton

Using coulomb's law , electric force of attraction between electron and proton  is given as

F = k q²/r²                          where k = constant = 9 x 10⁹

inserting the above values in the formula

F = (9 x 10⁹) (1.6 x 10⁻¹⁹)²/(2 x 10⁻¹⁰)²    

F = 5.76 x 10⁻⁹ N

as per newton's second law , acceleration is given as

a = F/m

hence acceleration of electron due to electrostatic force of attraction is given as

$a_{e}$ = F/m

inserting the values

$a_{e}$ = (5.76 x 10⁻⁹)/(9.1 x 10⁻³¹)

$a_{e}$ = 6.33 x 10²¹ m/s²

similarly

acceleration of proton due to electrostatic force of attraction is given as

$a_{p}$ = F/M

inserting the values

$a_{p}$ = (5.76 x 10⁻⁹)/( 1.67 x 10⁻²⁷)

$a_{p}$ = 3.45 x 10¹⁸ m/s²