In a mass spectrometer used for analyzing ion therapy beams, a beam of singly-charged Carbon-12 ions (12C+^{12}\text{C}^+12C+) enters a velocity selector. The selector contains a uniform magnetic field of flux density B=0.450 TB = 0.450\text{ T}B=0.450 T and a perpendicular uniform electric field of strength EEE.
Calculate the electric field strength EEE required so that the Carbon-12 ions travelling at a speed of 3.20×105 m s−13.20 \times 10^5\text{ m s}^{-1}3.20×105 m s−1 pass through the selector undeflected.
After emerging from the selector through the entry aperture, the ions enter a deflection chamber containing only the uniform magnetic field of flux density B=0.450 TB = 0.450\text{ T}B=0.450 T. Show that the radius rrr of this circular path is given by:
r=mvBqr = \frac{mv}{Bq}r=Bqmv
Calculate the distance between the entry aperture and the point on the detector surface where these 12C+^{12}\text{C}^+12C+ ions strike.
Given values:
- Charge of a Carbon-12 ion, q=1.60×10−19 Cq = 1.60 \times 10^{-19}\text{ C}q=1.60×10−19 C
- Mass of a 12C+^{12}\text{C}^+12C+ ion, m=1.99×10−26 kgm = 1.99 \times 10^{-26}\text{ kg}m=1.99×10−26 kg