Respuesta :
Answer:
a) [tex]4.1\times 10^{6} \frac{m}{s}[/tex]
b) [tex]9.2\times 10^{-8} s[/tex]
c) [tex]5.6\times 10^{-14} J[/tex]
d) 175000 volts
Explanation:
a)
[tex]q[/tex] = magnitude of charge on the alpha particle = 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C
[tex]m[/tex] = mass of alpha particle = 4 x 1.67 x 10⁻²⁷ kg = 6.68 x 10⁻²⁷ kg
[tex]r[/tex] = radius of circular path = 5.99 cm = 0.0599 m
[tex]B[/tex] = magnitude of magnetic field = 1.43 T
[tex]v[/tex] = speed of the particle
Radius of circular path is given as
[tex]r = \frac{mv}{qB}[/tex]
[tex]0.0599 = \frac{(6.68\times 10^{-27})v}{(3.2\times 10^{-19})(1.43)}[/tex]
[tex]v = 4.1\times 10^{6} \frac{m}{s}[/tex]
b)
Time period is given as
[tex]T = \frac{2\pi m}{qB}[/tex]
[tex]T = \frac{2(3.14)(6.68\times 10^{-27})}{(3.2\times 10^{-19})(1.43)}[/tex]
[tex]T = 9.2\times 10^{-8} s[/tex]
c)
Kinetic energy is given as
[tex]K = (0.5)mv^{2}[/tex]
[tex]K = (0.5)(6.68\times 10^{-27})(4.1\times 10^{6})^{2}[/tex]
[tex]K = 5.6\times 10^{-14} J[/tex]
d)
ΔV = potential difference
Using conservation of energy
q ΔV = K
(3.2 x 10⁻¹⁹) ΔV = 5.6 x 10⁻¹⁴
ΔV = 175000 volts
