Question

A proton moves along the x-axis where some arrangement of charges has produced the potential V(x)=Vosin(2πx/λ),where Vo = 3800 V and λ = 1.0 mm. a) What minimum speed must the proton have at x = 0 to move down the axis without being reflected? b) What is the maximum speed reached by a proton that at x = 0 has the speed you calculated in part A?

Answer 1

Answer with Explanation:

We are given that

`V(x)=Vsin((2\pi x)/(\lambda))`

`V_0=3800 V,\lambda=1.0 mm`

a.At x=0

V(0)=0

Maximum potential,
`V_(max)=V_0=3800 V`

The proton must have minimum speed when V(x) is maximum

Initial kinetic energy =
`q\Delta V`

`(1)/(2)mv^2=q(V_(max)-V(0))=q(3800-0)=3800q`

q=
`1.6* 10^(-19) C`

`m=1.67* 10^(-27) Kg`

`v^2=(2* 3800 q)/(m)=(3800* 2* 1.6* 10^(-19))/(1.67* 10^(-27))`

`v=\sqrt{(2* 3800 q)/(m)}=\sqrt{(3800* 2* 1.6* 10^(-19))/(1.67* 10^(-27))}`

`v=8.5* 10^5 m/s`

b.The maximum speed reached by a proton when V(x) is minimum

`V_(min)=-3800 V`

`v=\sqrt{(2q(V_(max)-V_(min))/(m)}`

`v=\sqrt{(2* 1.6* 10^(-19)(3800-(-3800)))/(1.67* 10^(-27))`

`v=1.21* 10^6 m/s`