Question

Suppose the electric field between the electric plates in the mass spectrometer of Fig. 20-38 is 2.38 ? 104 V/m and the magnetic fields B = B' = 0.66 T. The source contains carbon isotopes of mass numbers 12, 13, and 14 from a long-dead piece of a tree. (To estimate atomic masses, multiply by 1.67 ? 10-27 kg.)

A) How far apart are the lines formed by the singly charged ions of each type on the photographic film?
mm
(b) What if the ions were doubly charged?
mm

Answer 1

a)

E = electric field = 2.38 x 10⁴ N/C

B = magnetic field = 0.66 T

v = speed

speed is given as

v = E/B = (2.38 x 10⁴)/0.66 = 3.6 x 10⁴ m/s

m₁ = mass = 12 x 1.67 x 10^-27 kg

q = charge = 1.6 x 10^-19 C

B = magnetic field = 0.66 T

r₁ = m₁ v/(q₁ B) = (12 x 1.67 x 10^-27) (3.6 x 10⁴)/((1.6 x 10^-19 ) (0.66)) = 0.00683 m = 6.83 mm

m₂ = mass = 13 x 1.67 x 10^-27 kg

r₂ = m₂ v/(q₂ B) = (13 x 1.67 x 10^-27) (3.6 x 10⁴)/((1.6 x 10^-19 ) (0.66)) = 0.0074 m = 7.4 mm

m₃ = mass = 14 x 1.67 x 10^-27 kg

r₃ = m₃ v/(q₃ B) = (14 x 1.67 x 10^-27) (3.6 x 10⁴)/((1.6 x 10^-19 ) (0.66)) = 0.00797 m = 7.97 mm

sepration between lines r₂ - r₁ = 7.4 - 6.83 = 0.57 m

sepration between lines r₃ - r₂ = 7.97 - 7.4 = 0.57 m

b)

E = electric field = 2.38 x 10⁴ N/C

B = magnetic field = 0.66 T

v = speed

speed is given as

v = E/B = (2.38 x 10⁴)/0.66 = 3.6 x 10⁴ m/s

m₁ = mass = 12 x 1.67 x 10^-27 kg

q = charge = 2 x 1.6 x 10^-19 C

B = magnetic field = 0.66 T

r₁ = m₁ v/(q₁ B) = (12 x 1.67 x 10^-27) (3.6 x 10⁴)/((2 x 1.6 x 10^-19 ) (0.66)) = 3.4 mm

m₂ = mass = 13 x 1.67 x 10^-27 kg

r₂ = m₂ v/(q₂ B) = (13 x 1.67 x 10^-27) (3.6 x 10⁴)/((2 x 1.6 x 10^-19 ) (0.66)) = 3.7 mm

m₃ = mass = 14 x 1.67 x 10^-27 kg

r₃ = m₃ v/(q₃ B) = (14 x 1.67 x 10^-27) (3.6 x 10⁴)/((2 x 1.6 x 10^-19 ) (0.66)) = 3.985 mm

sepration between lines r₂ - r₁ = - (3.42 - 3.7) = 0.28 m

sepration between lines r₃ - r₂ = 3.985 - 3.7 = 0.285 m