Answer:
a) [tex] v = 2.36 \cdot 10^{7} m/s [/tex]
b) [tex] B = 3.80 \cdot 10^{-4} T [/tex]
c) [tex] f = 1.06 \cdot 10^{7} Hz [/tex]
d) [tex] T = 9.43 \cdot 10^{-8} s [/tex]
Explanation:
a) We can find the electron's speed by knowing the kinetic energy:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
K: is the kinetic energy = 1.59 keV
m: is the electron's mass = 9.11x10⁻³¹ kg
v: is the speed =?
[tex] v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2*1.59 \cdot 10^{3} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{9.11 \cdot 10^{-31} kg}} = 2.36 \cdot 10^{7} m/s [/tex]
b) The electron's speed can be found by using Lorentz's equation:
[tex] F = q(v\times B) = qvBsin(\theta) [/tex] (1)
Where:
F: is the magnetic force
q: is the electron's charge = 1.6x10⁻¹⁹ C
θ: is the angle between the speed of the electron and the magnetic field = 90°
The magnetic force is also equal to:
[tex] F = ma_{c} = m\frac{v^{2}}{r} [/tex] (2)
By equating equation (2) with (1) and by solving for B, we have:
[tex] B = \frac{mv}{rq} = \frac{9.11 \cdot 10^{-31} kg*2.36 \cdot 10^{7} m/s}{0.354 m*1.6 \cdot 10^{-19} C} = 3.80 \cdot 10^{-4} T [/tex]
c) The circling frequency is:
[tex] f = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{v}{2\pi r} [/tex]
Where:
T: is the period = 2π/ω
ω: is the angular speed = v/r
[tex] f = \frac{v}{2\pi r} = \frac{2.36 \cdot 10^{7} m/s}{2\pi*0.354 m} = 1.06 \cdot 10^{7} Hz [/tex]
d) The period of the motion is:
[tex] T = \frac{1}{f} = \frac{1}{1.06 \cdot 10^{7} Hz} = 9.43 \cdot 10^{-8} s [/tex]
I hope it helps you!
