008 (part 1 of 3) 10.0 points An alpha particle has a mass of 6.06 × 10 27 kg an

Question

008 (part 1 of 3) 10.0 points An alpha particle has a mass of 6.06 × 10 27 kg and bears a double elemen- tary positive charge. Such a particle is ob- served to move through a 4.5T magnetic field along a circular path of radius 0.18 r ular path of radius 0.18 m The charge on a proton is 1 .60218 x 10-19 C What speed does it have? Answer in units of m/s 009 (part 2 of 3) 10.0 points is its kinetic energy? What i Answer in units of J 010 (part 3 of 3) 10.0 points What potential difference in MV would be required to give it this kinetic energy? Answer in units of MV.

Explanation / Answer

given

mass = 6.06*10^(-27) kg
q = 1.60*10^(-19) C
R = 0.18 m
B = 4.5 T

the magnetic force
Fm = q*v*B    provides the centripetal force

So
q*v*B = ( m*v^2 )/R

q*B = m*v/R

part (1)

v = q*B*R / (m)

v = (2*1.60*10^(-19)) * (4.5) * (0.18) / (6.06*10^(-27))

v = 4.27 *10^7 m/s

v = 4.27 * 10^7 m/s

part(2)

kinetic energy Ek is

Ek = 1/2 * m * (v^2)

Ek = 1/2 *(6.06*10^(-27)) * ((4.27*10^7)^2)

Ek = 5.54*10^(-12) J

part(3)

Potential difference V is
V = (Ek) / (q)

V = (5.54*10^(-12)) / (2*1.60 * 10^(-19))

V = 17.31*10^(6) V

V = 17.31 MV

answer

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