- #1
lpettigrew
- 115
- 10
- Homework Statement
- A particle accelerator accelerates bunches of protons, each containing 115 billion particles, to an energy of 6.5 TeV (1 eV is the energy of one proton or electron that is accelerated by a potential of 1 Volt).
i. What is the electrical potential, in Volts, of each particle?
ii. If the proton charge is 1.6 * 10^(-19) C, what is the energy, in Joules, of each bunch?
iii. If one bunch reaches the collision point every 25 ns, what is the average electric current, in Amps, due to these arriving bunches?
- Relevant Equations
- W=V*Q
I=Q/t
Hello, I have answered the question below but would like some advice on whether I can improve my answer or if anyone is able to check whether I have made any mistakes ?
i. 1 V = 1eV in a 1:1 relationship, therefore;
6.5 TeV = 6.5 TV = 6.5 *10^12V
ii. E=W
W=V * Q
Q=number of particles * charge on each
Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8
W=(6.5 *10^12)*(1.84*10^-8)
W=119,749.5 J
iii. Find how many bunches reach the collision point in one second;
1 second = 1000000000 ns
1000000000ns/25ns= 40000000 = 4.0*10^7 bunches per second
The charge on a bunch of protons was earlier shown as 1.8423*10^-8 C.
So Q= ( 4.0 ^ 10^7)*(1.8423*10^-8)=0.73692 C
I=Q/t
I=0.73692/2.5*10^-8 s
I=29476800 A
I think that I have taken a wrong turn on the last question and have confused myself a little confessedly
i. 1 V = 1eV in a 1:1 relationship, therefore;
6.5 TeV = 6.5 TV = 6.5 *10^12V
ii. E=W
W=V * Q
Q=number of particles * charge on each
Q=(115*10^9)*(1.602*10^-19)=1.8423*10^-8
W=(6.5 *10^12)*(1.84*10^-8)
W=119,749.5 J
iii. Find how many bunches reach the collision point in one second;
1 second = 1000000000 ns
1000000000ns/25ns= 40000000 = 4.0*10^7 bunches per second
The charge on a bunch of protons was earlier shown as 1.8423*10^-8 C.
So Q= ( 4.0 ^ 10^7)*(1.8423*10^-8)=0.73692 C
I=Q/t
I=0.73692/2.5*10^-8 s
I=29476800 A
I think that I have taken a wrong turn on the last question and have confused myself a little confessedly