Question

Consider an electron and a proton that are initially at rest and are separated by 2.00 nm. Neglecting any motion of the much more massive proton, what is the minimum (a) kinetic energy and (b) speed that the electron must be projected at so it reaches a point a distance of 12.0 nm from the proton? Assume the electron’s velocity is directed radially away from the proton. (c) How far will the electron travel away from the proton if it has twice that initial kinetic energy?

Answer 1

Answer:

a)$W = 57.6\times10^{-19} J$

b) $v = 3.556\times10^6 m/s$

c) displacment = 400nm

Explanation:

Given that

charge of the electron and proton $q =1.6\times10^{-19} C$

the distance between electron and proton = 2.0 nm

The force between the proton and electron

$F = kq^/r^2$

Here $k = 9\times10^9 N.m^2/C^2$

$F = (9\times10^9)(1.6\times10^-19)(1.6\times10^-19)/(2.0\times10^-9)^2 \\= 5.76\times10^-11N$

The work done in displacing the electron from 2.0 nm to 12 nm

is the minimum kinetic energy

W = F .s

$W = (5.76\times10^{-11}N)(10\times10^{-9} m)\\W = 57.6\times10^{-19} J$

(b)

If this energy is the kinetic energy of the electron

$(1/2) m v^2 \\= 57.6\times10^{-19} J$

Here m is the mass of the electron =$9.0\times10^{-31} kg$

The speed of the electron $v = 3.556\times10^6 m/s$

(c)

If the kinetic energy is twice the initial kinetic energy

2*(initial kinetic energy) = F*displacement

displacement = 2*(initial kinetic energy)/F

Displacement = $2(57.6\times10^{-19} J)/(5.76\time10^{-11}N)$

displacement = 400 nm