Questions on Rutherford scattering and the nuclear model

Question 1

Describe the main observations of the Rutherford alpha-particle scattering experiment.

Accepted Answer

Most alpha particles passed through the thin gold foil with little or no deflection (1); a small fraction were deflected through large angles, with a very few deflected back through more than 90° (1).

Question 2

In the Rutherford scattering experiment, most alpha particles passed straight through the gold foil, while a tiny fraction were deflected through large angles. State the conclusions about the structure of the atom drawn from these observations.

Accepted Answer

The atom is mostly empty space, since most particles pass straight through (1); the positive charge and nearly all the mass are concentrated in a very small central nucleus, which repels the few particles that pass close to it (1).

Question 3

State what is meant by background radiation and give two of its origins.

Accepted Answer

Ionising radiation that is always present, from sources other than the source under investigation (1). Any two origins: radon gas from the ground, rocks and building materials, cosmic rays, radioactive traces in food, medical procedures (1).

Question 4

State the nature of an alpha particle and of gamma radiation.

Accepted Answer

An alpha particle is a helium nucleus: two protons and two neutrons (1). Gamma radiation is a high-frequency electromagnetic wave (a photon) (1).

Question 5

State a material that will absorb each of alpha, beta and gamma radiation.

Accepted Answer

Alpha: a sheet of paper (or a few cm of air) (1). Beta: a few mm of aluminium (1). Gamma: several cm of lead reduces the intensity (1).

Question 6

A source emits three types of radiation. One is stopped by paper. A second passes through paper but is stopped by a few millimetres of aluminium. A third passes through both and is only reduced by lead. Identify each radiation.

Accepted Answer

Stopped by paper: alpha (1). Stopped by aluminium but not paper: beta (1). Only reduced by lead: gamma (1).

Question 7

A gamma source produces an intensity of 800 arbitrary units at a distance of 1.0 m. Assuming an inverse-square law, calculate the intensity at 2.0 m.

Accepted Answer

I ∝ 1/r², so I = 800 × (1.0/2.0)² (1) I = 200 units (1)

Question 8

A detector reads 1600 counts per second at a distance of 0.50 m from a gamma source. Assuming an inverse-square law, calculate the expected reading at 2.0 m.

Accepted Answer

I = 1600 × (0.50/2.0)² = 1600 × 1/16 (1) I = 100 counts per second (1)

Question 9

Explain why the great majority of alpha particles pass through a thin gold foil almost undeflected.

Accepted Answer

The atom is mostly empty space (1); the nucleus occupies only a tiny fraction of the atom's volume (1); most alpha particles pass far from any nucleus, experience very little electrostatic repulsion, and so are barely deflected (1).

Question 10

Before Rutherford's experiment, the atom was modelled as a sphere of positive charge with electrons embedded in it. Explain why the deflection of some alpha particles through more than 90° could not be explained by this model.

Accepted Answer

With the positive charge spread through the whole atom, the electric field acting on an alpha particle would everywhere be weak (1); each alpha particle would therefore be deflected only through a very small angle (1); deflection through more than 90° needs a very strong repulsive force, which requires the positive charge (and most of the mass) to be concentrated in a very small volume, not spread out (1).

Question 11

A detector placed 150 mm from a sealed gamma source records a corrected count rate of 3600 counts per second. Assuming the intensity obeys an inverse-square law, determine the distance from the source at which the corrected count rate falls to 400 counts per second. Give your answer in metres.

Accepted Answer

The count rate must fall by a factor of 3600/400 = 9 (1) I ∝ 1/x², so the distance must increase by a factor of √9 = 3 (1) x = 3 × 150 mm = 0.45 m (1)

Question 12

A domestic smoke detector contains a source that emits only alpha particles. Explain why the source presents very little hazard while sealed in the detector, but would be dangerous if the source material were swallowed.

Accepted Answer

Alpha particles are absorbed by a few centimetres of air (or by the detector casing, or by the dead outer layer of skin), so from outside the body the radiation cannot reach living tissue (1). If the material were swallowed, alpha particles would be emitted directly into living tissue (1); alpha is the most strongly ionising radiation, so all of its energy would be absorbed in a very small volume of tissue, causing serious damage (1).

Question 13

With a gamma source in place, a detector at 0.10 m records 520 counts per second, including a background count of 20 counts per second. Use the inverse-square law to predict the measured count rate at 0.20 m.

Accepted Answer

Corrected count rate at 0.10 m = 520 − 20 = 500 counts per second (1) At 0.20 m: 500 × (0.10/0.20)² (1) = 125 counts per second (1) Measured rate = 125 + 20 = 145 counts per second (1)

Question 14

Explain why a few alpha particles in the scattering experiment were deflected through very large angles, and state which particles are deflected the most.

Accepted Answer

A large deflection occurs when an alpha particle passes very close to a nucleus (1); the concentrated positive charge exerts a strong electrostatic repulsion on the (positive) alpha particle (1); particles heading almost directly at a nucleus approach closest and are deflected most, some through more than 90° (1).

Question 15

Compare the ionising power and the penetrating power of alpha, beta and gamma radiation.

Accepted Answer

Alpha: most strongly ionising, least penetrating, stopped by paper (1). Beta: moderate ionising and penetrating power, stopped by a few mm of aluminium (1). Gamma: least ionising, most penetrating, only reduced by thick lead (1).

Question 16

A factory rolls aluminium foil about 0.1 mm thick. The thickness is monitored continuously by a radioactive source above the moving foil and a detector below it. Four sources are available. W: alpha emitter, half-life 432 years; X: beta emitter, half-life 29 years; Y: beta emitter, half-life 15 hours; Z: gamma emitter, half-life 5.3 years. Suggest, with reasons, which source the factory should choose.

Accepted Answer

Alpha (W) is absorbed completely by even very thin foil, so the count rate would be near zero whatever the thickness (1). Gamma (Z) passes through thin foil almost unaffected, so the count rate would barely respond to a change in thickness (1). A beta source is suitable because its absorption changes measurably over thicknesses of about 0.1 mm (1). X is chosen rather than Y: its long half-life means the activity is effectively constant, so a change in count rate indicates a change in thickness rather than decay of the source (1).

Question 17

A student measures the count rate near a sealed gamma source. With the detector 0.20 m from the source the measured count rate is 340 counts per second; at 0.40 m it is 100 counts per second. Both readings include a constant background count rate. Assuming the gamma intensity obeys an inverse-square law, determine the background count rate.

Accepted Answer

Doubling the distance reduces the corrected (source-only) count rate to one quarter (1)
So 340 − B = 4 × (100 − B) (1)
340 − B = 400 − 4B, so 3B = 60 and B = 20 counts per second (1)
Check: the corrected rates are 320 and 80 counts per second, a factor of 4 apart as the law requires (1)

Question 18

Describe an experiment to test whether the intensity of the gamma radiation from a small sealed source obeys an inverse-square law. Include the measurements taken, how the data are analysed, and the precautions taken to keep the experimenter's dose low.

Accepted Answer

With the source locked away, measure the background count rate over a long period (1). Place the source a measured distance x from a Geiger-Müller tube, using a metre rule, and record the count rate; repeat for a range of distances, counting for long enough (or repeating) to reduce the random uncertainty of the counts (1). Subtract the background count rate from every reading to give the corrected count rate C (1). The rule measures from the outside of the casings, but the source sits inside its holder and the tube's sensitive volume lies behind its window, so the true separation is x + x₀ for some unknown constant offset x₀ (1). With C = k/(x + x₀)², 1/√C = (x + x₀)/√k, so plot 1/√C against x: a straight line confirms the inverse-square law, it need not pass through the origin, and its intercept divided by its gradient gives the offset x₀ (1). Precautions: handle the source with long tongs, stand well back, keep the exposure time short, and return the source to its lead-lined store immediately after use (1).