Problem 19.40

19.40. a)
We first need to come up with an expression for the acceleration for the accleration of the charged particles. The acceleration of the proton is given by
$a_{p} = \frac{qE}{m_{proton}}$
$a_{p} = 6.13 \times 10^{10} m/s^{2}$
Now, for the electron, the only thing that changes (other than the sign of the acceleration) is the mass. So, for the electron,
$a_{e} = \frac{qE}{m_{electron}}$
$a_{e} = 1.12 \times 10^{14} m/s^{2} ~=~ 1836 a_{p}$
The distance that the proton travels in a time t is given by
$x_{p} = \frac{1}{2} a_{p} t^{2}$
and similarly, for the electron,
$x_{e} = \frac{1}{2} a_{e} t^{2}$
When they meet, the total distance is 4 cm. So,
$.04 = \frac{1}{2} a_{p} t^{2} + \frac{1}{2} a_{e} t^{2}$
Using the fact found earlier that the acceleration of the electron is 1836 times that of the acceleration of the proton, we find that
$(.04) \frac{1}{1837} = \frac{1}{2} a_{p} t^{2} = x_{p}$
or, $x_{p} = 21.8 \times 10^{-6} m$

b)
An identical approach is used here.
$a_{Na} = \frac{eE}{22.99 amu}$
$a_{Cl} = \frac{eE}{35.45 amu}$
So, the acceleration of the sodium ion is 1.54 times greater than that of the chlorine ion. Solving for the distance travelled by the sodium ion when they collide we find
$x_{Na} = \frac{4 cm}{1 + 1.54} ~=~ 1.57 cm$