Question

Item 4 part a if a proton and an electron are released when they are 4.50×10−10 m apart (typical atomic distances), find the initial acceleration of each of them.

Answer 1

Acceleration is defined as the rate of change of velocity with respect to time.

Formulas of force are given by:

$F=ma$   (1)

where,

F = force

m = mass

a = acceleration

$F=\frac{kq_{1}q_{2}}{r^{2}}$   (2)

where,

F = force

k = coulomb constant ($9\times 10^{9}Nm^{2}C^{-2}$)

r = distance between the charged particles

$q_{1}$ and $q_{2}$ are the signed magnitude of charges

Use the formula (2) for calculating the value of force, we get:

Substitute the value of $q_{1}$ and $q_{2}$, k and r to find the value of force.

F= $\frac{9\times 10^{9} Nm^{2}C^{-2}\times (1.6\times 10^{-19} C)^{2}}{(4.50\times 10^{-10})^{2}}$

= $1.1378\times 10^{-9} N$

Now, put the above value force in formula (1) to identify the initial acceleration.

$1.1378\times 10^{-9} N=ma$   (1)   (mass of electron  =$9.1 \times 10^{-31} kg$)

acceleration of electron = $\frac{1.1378\times 10^{-9} N}{9.1\times 10^{-31}kg}$

= $1.25\times 10^{21}N/kg$

And,

acceleration of proton =$\frac{1.1378\times 10^{-9} N}{mass of proton}$

mass of proton =$1.67 \times 10^{-27}$

Thus,

acceleration of proton = $\frac{1.1378\times 10^{-9} N}{1.67 \times 10^{-27} kg}$

=$6.81 \times 10^{17} N/kg$

Now, initial acceleration of electron and proton is  $1.25\times 10^{21}m/s^{2}$ and $6.81 \times 10^{17} m/s^{2}$ as 1 N = $kgm/s^{2}$.