Answer:
v = 61.09m/s
Explanation:
In order to calculate the speed of the electron when it is 3.00cm from the proton, you first calculate the acceleration of the electron, produced by the electric force between the electron and the proton. By using the second Newton law you have:
[tex]F=ma=k\frac{q^2}{r^2}[/tex] (1)
m: mass of the electron = 9.1*10^-31kg
q: charge of electron and proton = 1.6*10^-19C
r: distance between electron and proton = 9.00cm = 0.09m
k: Coulomb's constant = 8.98*10^9Nm2/C^2
You solve the equation (1) for a, and replace the values of the other parameters:
[tex]a=\frac{kq^2}{mr^2}=\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)^2}{(9.1*10^{-31}kg)(0.09m)^2}=3.11*10^4\frac{m}{s^2}[/tex]
Next, you use the following formula to calculate the final speed of the electron:
[tex]v^2=v_o^2+2ax[/tex] (2)
vo: initial speed of the electron = 0m/s
a: acceleration = 3.11*10^4m/s^2
x: distance traveled by the electron
When the electron is at 3.00cm from the proton the electron has traveled a distance of 9.00cm - 3.00cm = 6.00cm = 0.06m = x
You replace the values of the parameters in the equation (2):
[tex]v=\sqrt{2ax}=\sqrt{2(3.11*10^4m/s)(0.06m)}=61.09\frac{m}{s}[/tex]
The speed of the electron is 61.09m/s
