IB · D · extended response
A velocity selector followed by a magnetic analyser
Fields · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Mass analyser calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Orbit radius / m
Horizontal axis (x): Ion mass / 10⁻²⁷ kg
Read the exact data as a table
| Ion mass / 10⁻²⁷ kg | Orbit radius / m |
|---|---|
| 2 | 0.0125 |
| 4 | 0.025 |
| 6 | 0.0375 |
| 8 | 0.05 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Positive singly charged ions first cross perpendicular uniform electric and magnetic fields E = 2.0×10⁴ N C⁻¹ and Bselector = 0.10 T, oriented so their forces oppose. Undeflected ions then enter a separate perpendicular analyser field 0.20 T. Charge is e = 1.60×10⁻¹⁹ C. The supplied calibration gives analyser radius against mass for ions at the selected speed. Supplied data (Ion mass / 10⁻²⁷ kg; Orbit radius / m): (2, 0.0125), (4, 0.025), (6, 0.0375), (8, 0.05).
- (a)
Find selected speed and the mass of an ion following a 0.0375 m analyser radius. Explain the origin of the circular path.
3 marks - (b)
Predict the effect of a doubly charged ion of the same mass at the selected speed, and describe the analyser's sign discrimination.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find selected speed and the mass of an ion following a 0.0375 m analyser radius. Explain the origin of the circular path.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
In the selector, qE = qvB gives v = E/B = 2.0×10⁵ m s⁻¹; charge cancels, so this stage selects speed rather than mass.
- 2
In the analyser qvB = mv²/r, so m = qBr/v = (1.60×10⁻¹⁹)(0.20)(0.0375)/(2.0×10⁵) = 6.0×10⁻²⁷ kg.
- 3
The magnetic force is perpendicular to velocity, changing its direction and providing the circular resultant without doing work on the ion.
(b)
Predict the effect of a doubly charged ion of the same mass at the selected speed, and describe the analyser's sign discrimination.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The selected speed E/B does not depend on charge magnitude when both selector forces scale with q.
- 2
In the analyser r = mv/(|q|B), so doubling charge halves radius at fixed mass, speed and field. The apparatus measures mass-to-charge ratio, not mass independently.
- 3
Changing charge sign reverses the magnetic force direction. Field direction and incoming velocity must be specified to name the clockwise or counterclockwise sense; radius alone gives no sign.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- In the selector, qE = qvB gives v = E/B = 2.0×10⁵ m s⁻¹; charge cancels, so this stage selects speed rather than mass.
- In the analyser qvB = mv²/r, so m = qBr/v = (1.60×10⁻¹⁹)(0.20)(0.0375)/(2.0×10⁵) = 6.0×10⁻²⁷ kg.
- The magnetic force is perpendicular to velocity, changing its direction and providing the circular resultant without doing work on the ion.
- The selected speed E/B does not depend on charge magnitude when both selector forces scale with q.
- In the analyser r = mv/(|q|B), so doubling charge halves radius at fixed mass, speed and field. The apparatus measures mass-to-charge ratio, not mass independently.
- Changing charge sign reverses the magnetic force direction. Field direction and incoming velocity must be specified to name the clockwise or counterclockwise sense; radius alone gives no sign.
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