The magnitude of the change in potential energy is equal to the kinetic energy,
[tex]\left | \Delta U \right |= K[/tex]
or
[tex]qV=\frac{1}{2} mv^{2}[/tex]
Here, V is potential difference, q is charge, m is mass and v is velocity.
We can also write,
[tex]v=\sqrt{\frac{2qV}{m} }[/tex]
Given [tex]V=380 V[/tex]
Substituting this value with mass of proton, [tex]m=1.672\times10^{-27} kg[/tex] and charge of proton,[tex]q= 1.6\times10^{-19} C[/tex] we get
[tex]v=\sqrt{\frac{2\times1.6\times10^{-19}C \times 380V}{1.672\times10^{-27} kg} }\\\\v= 727.27\times10^{8} m/s =7.27\times10^{6}m/s[/tex]
Therefore, the speed of proton is [tex]7.27\times10^{6}m/s[/tex].
Answer:
The speed of a proton after it accelerates from rest is 2.698x10⁵ m/s
Explanation:
Please look at the solution in the attached Word file
