Answer:
9.17 x 10^-10 m
Explanation:
q = 1.6 x 10^-19 C
B = 4.55 x 10^4 T
m = 9.1 x 10^-31 kg
K = 22.5 eV = 22.5 x 1.6 x 10^-19 J = 3.6 x 10^-18 J
θ = 65.5°
Use the formula for the kinetic energy
Let r be the radius of circular path and v be th evelocity
[tex]K = \frac{B^{2}q^{2}r^{2}}{2m}[/tex]
By substituting the values
[tex]3.6 \times 10^{-18} = \frac{\left ( 4.55 \times 10^{4} \right )^{2}(1.6\times 10^{-19} \right )^{2}r^{2}}{2\times 9.1\times 10^{-31}[/tex]
r = 3.52 x 10^-10 m
Use the formula for the speed
[tex]v =\frac{Bqr}{m}[/tex]
[tex]v =\frac{4.55 \times 10^{4}\times 1.6\times 10^{-19}\times 3.52 \times 10^{-10}}{9.1\times 10^{-31}}[/tex]
v = 2.82 x 10^6 m/s
The pitch of the helix = [tex]v Cos\theta \times \frac {2\pi r}{v}[/tex]
= [tex]2\pi rCos\theta =2 \times 3.14 \times 3.52 \times 10^{-10} \ Cos65.5[/tex]
= 9.17 x 10^-10 m
