Answer:
a. i. 8.447 × 10⁻³ T ii. 27.14 cm
b. i. 2.14 cm ii. It is easily detectable.
Explanation:
a.
i. What strength of magnetic field is required?
Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have
F = F'
Bqv = mv²/r
B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m
So,
B = mv/rq
B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)
B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)
B = 0.8447 × 10⁻² kg/sC)
B = 8.447 × 10⁻³ T
(ii) What is the diameter of the 13C semicircle?
Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have
F = F'
Bq'v = mv²/r'
r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and = d/2 = 12.5 cm = 12.5 × 10⁻² m
So, r' = mv/Be
r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)
r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)
r' = 1.357 × 10⁻¹ kgm/TC)
r' = 0.1357 m
r' = 13.57 cm
Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm
b.
i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?
Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm
ii. Is this distance large enough to be easily observed?
This distance of 2.14 cm easily detectable since it is in the centimeter range.
