Magnetic field: the Banebridge velocity selector

[Karol]

Homework Statement

The electric field intensity between the plates of the Banebridge velocity selector (I am not a native English speaker …) is 120 [Volt/cm] and the magnetic field magnitude of both fields is 0.6 [Weber/m²].
A singly charged Neon ion beam makes a radius of 7.28 [cm]. What is it's mass number.
The result must be 21, 10 times smaller than mine.

Homework Equations

Velocity of beam: V=\frac{E}{B}
radius of the beam: R[m]=\frac{mV}{qB}
mass number=mass/proton mass: N=\frac{m}{m_{p}}

The Attempt at a Solution

120[Volt/cm]=12,000[Volt/m]

V=\frac{E}{B}=\frac{12,000}{0.6}=20,000[m/sec]

R=0.728=\frac{mV}{qB}=\frac{m*20,000}{1.602\ \times\ 10^{-19}*0.6}\Rightarrow m=3.5\ \times\ 10^{-25}

N=\frac{m}{m_{p}}=\frac{3.5\ \times\ 10^{-25}}{1.672\ \times\ 10^{-27}}=209

 

[thebigstar25]

what is your questions exactly?

 

[Karol]

My result is 10 times bigger, according to the book and the periodic table of elements, it should come out 20.9

 

[thebigstar25]

I can't see what's wrong with your solution , it seems okay ..

 

[Karol]

Thanks for your effort, Ihope I'll get another answer before I decide there was a mistake in the book. By the way, it's taken from a translation to Sears & Zemansky, 1964...