Respuesta :
Answer:
Radius of the path is [tex]163.16\times 10^{-6}m[/tex]
Explanation:
It is given that gold is doubly ionized
Charge [tex]q=2e=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}C[/tex]
Mass of the each ion = [tex]=3.27\times 10^{-25}kg[/tex]
So total mass m = [tex]=2\times 3.27\times 10^{-25}=6.54\times 10^{-25}kg[/tex]
Potential difference V = 1.9 KV
Magnetic field B = 0.540 T
We know that [tex]\frac{1}{2}mv^2=qV[/tex]
[tex]\frac{1}{2}\times 6.54\times 10^{-25}\times v^2=3.2\times 10^{-19}\times 1900[/tex]
[tex]v=43.11\times 10^3m/sec[/tex]
We have to find the radius of the path
We know that radius of the path is given by
[tex]r=\frac{mv}{qB}=\frac{6.54\times 10^{-25}\times 43.11\times 10^3}{3.2\times 10^{-16}\times 0.540}=163.16\times 10^{-6}m[/tex]
