A mass spectrometer, sketched ,is a device used to separate different ions. Such ions with a well-defined velocity vo enter through a slit into a region of uniform magnetic field B, where they follow a semicircular path until they strike the detector slit above the entry slit. The distance between the entry and the detector slits is d = 1.20 m. Chlorine ions of mass 35 amu (1 amu equals 1.66x10-27 kg), carrying a charge of +1e, enter the spectrometer with initial speed of vo = 8.70E5 meters/sec. What value of B is required for these ions to hit the detector?
Equally charged chlorine ions of mass 37 amu now enter the spectrometer. How close to the detector slit will they impact?

Explanation:

From the given data , it appears that , the ion particles are moving on a circular path of radius 1.2 /2  =0.6 m in the magnetic field .

In the circular path , centripetal force is  provided by magnetic force on ions

m v² / R = Bq v

B = mv / q R

= 35 x 1.66 x 10⁻²⁷x 8.7 x 10⁵ / 1.6 x 10⁻¹⁹ x .6

= 526.53 x 10⁻³ T .

If mass of the ion becomes 37 x 1.66 x 10⁻²⁷ , the radius of circular path will be changed

m v² / R = Bq v

R = mv / qB

37 x 1.66 x 10⁻²⁷ x 8.7 x 10⁵ / 1.6 x 10⁻¹⁹ x 526.53 x 10⁻³

= .634 m

difference = .634 - .6

.034 m

= 3.4 cm

The ion will hit 3.4 cm away from the detector .

that the claims meet official rules.

don't misinterpret the claims

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr