Singly ionized carbon atoms ( atoms with one electron removed) are sent through

Question

Singly ionized carbon atoms ( atoms with one electron removed) are sent through a mass spectrometer. The first stage is a velocity selector: they first travel through a region in which the electric field has a magnituse of 1.25 x 10^4 N/C, vertically upward, and the magnetic field 0.0200 T, directed out of the page. At what speed must the ions travel in order that the electric and magnetic forces cancel each other?

If the carbon ions from the previous problem left the velocity selector and entered a region where the magnetic field had the same megnitude and direction as before, but with no electric field, the ions begin to move along circular paths. What is the mass of the ion if the radius of the circular path is:

a) 3.886m

b) 4.209m

c)4.533m

Explanation / Answer

here by using the formula

q * E = q * v * B

v = E / B

v = (1.25 * 10^4) / (0.02)

v = 6.25 * 10^5 m/s

so the speed of ion is 6.25 * 10^5 m/s

then

a )

at r = 3.886 m

here by using the formula

r = m * v / (q*B )

3.886 = m * 6.25 * 10^5 / ( 1.602 * 10^-19 * 0.02)

by solving this for m

the , m = 1.992 * 10^-26 kg

b)

at r = 4.209 m

here also we use the same formula

m = r * q * B / v

m = (4.209 * 1.602 * 10^-19 * 0.02) / (6.25 * 10^5)

m = 2.157 * 10^-26 kg

c)

r = 4.533 m

here also

m = (4.533 * 1.602 * 10^-19 * 0.02) / (6.25 * 10^5)

m = 2.32 * 10^-26 kg

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