Answer:
(a) v = 2.93x10⁶ m/s
(b) T = 1.43x10⁻⁷ s
(c) K = 1.79x10⁵ eV
(d) ΔV = 89.5 kV
Explanation:
(a) The speed of the alpha particle can be found using the magnetic force (Fm) and the centripetal force (Fc):
[tex]Fm = Fc[/tex]
[tex] qvB = \frac{mv^{2}}{r} [/tex]
[tex] v = \frac{qrB}{m} [/tex] (1)
Where:
q: is the charge on the alpha particle = 2e⁻ = 2*1.602x10⁻¹⁹ C = 3.20x10⁻¹⁹ C
r: is the radius of the circular path = 6.68 cm = 6.68x10⁻² m
B: is the magnetic field = 0.917 T
m: is the mass of the alpha particle = 4.00 u = 4*1.67x10⁻²⁷ kg = 6.68x10⁻²⁷kg
Hence, the speed of the alpha particle is:
[tex] v = \frac{qrB}{m} = \frac{3.20\cdot 10^{-19} C*6.68 \cdot 10^{-2} m*0.917 T}{6.68 \cdot 10^{-27} kg} = 2.93 \cdot 10^{6} m/s [/tex]
(b) The period (T) of revolution of the alpha particle is:
[tex]T = \frac{2\pi r}{v} = \frac{2\pi 0.0668 m}{2.93 \cdot 10^{6} m/s} = 1.43 \cdot 10^{-7} s[/tex]
(c) The kinetic energy (K) of the alpha particle is:
[tex]K = \frac{1}{2}mv^{2} = \frac{1}{2}6.68 \cdot 10^{-27} kg*(2.93 \cdot 10^{6} m/s)^{2} = 2.86 \cdot 10^{-14} J*\frac{1 eV}{1.60 \cdot 10^{-19} J} = 1.79 \cdot 10^{5} eV[/tex]
(d) The potential difference (ΔV) is:
[tex]\Delta V = \frac{K}{q} = \frac{1.79 \cdot 10^{5} eV}{2e} = 89.5 kV[/tex]
I hope it helps you!
