The diagram shows a portable beta-source tester used to check the magnetic coils in a teaching laboratory. A strontium-90 check source is kept in a protecti...
Question 1 Report
The diagram shows a portable beta-source tester used to check the magnetic coils in a teaching laboratory. A strontium-90 check source is kept in a protective cap until the test begins. Beta particles enter the evacuated box through a narrow opening. Current in the coils produces a magnetic field into the page, shown by crosses. The curved path is observed using a light-sensitive screen. A potential difference of 2500 V is applied across the accelerating section before the particles reach the magnetic field.
Fig. 1 shows the path of a beta particle. The particle enters from the left and curves downwards.
(a) What is the charge on a beta particle? [1]
(b) Describe the direction of the magnetic force on the particle as it first enters the field. [1]
(c) Explain why a beta particle with a lower velocity would follow a more tightly curved path in the same magnetic field. [3]
(d) Give two precautions used when handling the strontium-90 source. [2]
(e) Use the potential difference to calculate the energy gained by one beta particle, in J. [3]
Answer Details
(a) A beta particle has negative charge. [1]
(b) As it first enters the field, the magnetic force is downwards, as shown by the initial downward curvature of its path. [1]
(c) A moving charged particle experiences a magnetic force in a magnetic field. This force changes the direction of the particle, producing a curved path. A lower velocity means lower momentum, so the same magnetic force produces a smaller radius of curvature, giving a tighter curve. [3]
(d) Keep the source in its protective cap or shielded container when not in use, and use tongs or a remote holder. Minimising handling time, keeping it away from the body, and using a labelled secure container are also acceptable. [2]
(e) Use \(E=QV\):
\[E=(1.6\times10^{-19}\text{ C})(2500\text{ V})\]
\[E=4.0\times10^{-16}\text{ J}\]
The energy gained by one beta particle is \(4.0\times10^{-16}\text{ J}\). [3]
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