Answer :
Answer:
The velocity of the proton is 4.25 x 10³ m/s
Explanation:
Given;
speed of electron, Ve = 7.80 x 10⁶ m/s
magnetic field strength, B = 1.05 x 10⁻⁵ T
In cyclotron motion, the radius of a charged particle path in a magnetic field is given as:
[tex]r= \frac{mv}{qB}[/tex]
where;
m is the mass of the particle
v is the speed of the particle
q is the charge of the particle
B is the strength of the magnetic field
If both electron and proton travel in a path with same radius, our equation becomes;
[tex]r = \frac{M_eV_e}{qB} = \frac{M_pV_p}{qB} \\\\M_eV_e=M_pV_p\\\\V_p = \frac{M_eV_e}{M_p}[/tex]
Me is the mass of electron = 9.1 x 10⁻³¹ kg
Mp is the mass of proton = 1.67 x 10⁻²⁷ kg
[tex]V_p = \frac{M_eV_e}{M_p} = \frac{9.1*10^{-31}*7.8*10^6}{1.67*10^{-27}} = 4.2503 *10^3 \ m/s[/tex]
Therefore, the velocity of the proton is 4.25 x 10³ m/s
Answer:
Vp = 4.25×10³m/s
Explanation:
Given B = 1.05×10-⁵T, Ve = velocity of electron = 7.80×10⁶m/s.
Mass of electron me = 9.11×10-³¹kg, mp = 1.67×10-²⁷kg
Both particles move in a circular path of the same radius R so Re = Rp
R = mV/qB
The magnetic field is the same for both particles and the charge is also the same for both particles.
So R = meVe/ qB = mpVp/qB
Since the product qB is common to both sides,
meVe = mpVp
Which is basically a statement that both particles have the same momentum.
Vp = meVe/mp = (9.11×10-³¹×7.80×10⁶)/(1.67×10-²⁷)
Vp = 4254.97m/s = 4.25×10³m/s