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In their famous experiment, Rutherford's assistants, Geiger and Marsden, detected the scattered alpha particles using a zinc sulphide screen, which produced a tiny flash of light when struck by an alpha particle. If their screen had area \(1 \mathrm{mm}^{2}\) and was \(1 \mathrm{cm}\) from the target, what solid angle did it subtend?

Short Answer

Expert verified
The solid angle subtended is 0.01 steradians.

Step by step solution

01

Understand the Problem

We are asked to find the solid angle subtended by a screen of area \(1 \text{ mm}^2\) placed \(1 \text{ cm}\) from the target. The solid distance in question is essentially the cone spread from the target, capturing the area of the screen.
02

Convert Units

To perform calculations using consistent SI units, we first convert the area of the screen to square centimeters. Since \(1 \text{ mm}^2 = 0.01 \text{ cm}^2\), the area \(A\) is \(0.01 \text{ cm}^2\). The distance \(r\) from the target to the screen is already in centimeters: \(1 \text{ cm}\).
03

Apply the Solid Angle Formula

The solid angle \(\Omega\) subtended by a surface at a distance \(r\) is given by the formula \(\Omega = \frac{A}{r^2}\), where \(A\) is the area of the surface and \(r\) is the distance from the target. Substituting our values, we have: \[ \Omega = \frac{0.01}{1^2} = 0.01 \text{ steradians} \].
04

Conclusion

We find that the screen subtends a solid angle of \(0.01\) steradians from the target. This is a measure of how large the screen appears to the target in three-dimensional space.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rutherford's Experiment
Rutherford's experiment, which took place in the early 20th century, was pivotal in our understanding of the atomic structure. Ernest Rutherford, along with his colleagues Geiger and Marsden, conducted an experiment that involved firing alpha particles at a thin gold foil. They observed how these particles scattered, revealing unexpected outcomes.

Prior to this experiment, the prevalent model of the atom was J.J. Thomson's "plum pudding model," which suggested that atoms were composed of a diffused positive charge with electrons embedded like plums in a pudding. Rutherford's experiment overturned this model by showing that atoms have a tiny, dense nucleus.

The setup consisted of a source of alpha particles, a thin gold foil, and a zinc sulphide screen to detect scattered particles. The observation was that while many particles passed through the foil with little deflection, some were deflected at large angles, and a few even bounced back towards the source. This led to the revolutionary conclusion that a dense core (the nucleus) existed within the atom, with electrons orbiting this nucleus.
Alpha Particle Scattering
Alpha particle scattering was critical in unveiling the structure of the atom in Rutherford's experiment. Alpha particles are essentially helium nuclei, consisting of two protons and two neutrons. When Rutherford's team directed these positively charged particles at a gold foil, the scattering pattern provided insights into atomic structure.

There are several key observations from the scattering experiment:
  • Most alpha particles passed straight through the foil, indicating that atoms are mostly empty space.
  • Some alpha particles were deflected at small angles, suggesting that they were passing close to a positively charged center which exerted a repulsive force.
  • A very few bounced back, which could only occur if they encountered something very small and dense, i.e., the nucleus.
These observations laid the groundwork for the nuclear model of the atom, where a dense nucleus holds most of the atom's mass and positive charge, surrounded by electrons in the surrounding space.
Steradian
The concept of steradians is vital in understanding measurements involving solid angles, like those used in Rutherford's experiment. A steradian is the SI unit of solid angle and can be thought of as the three-dimensional equivalent of the radian, which measures planar angles.

In practical terms, it quantifies how much of a sphere's surface is covered or seen from a particular point. The formula for calculating the solid angle \( \Omega \) subtended by a surface is based on its area \( A \) and the square of the distance \( r \) from the observer: \[ \Omega = \frac{A}{r^2} \].

In the context of Rutherford's experiment, the screen used for detecting alpha particles subtended a certain solid angle from the gold foil's perspective. Calculating this solid angle helps in understanding how much of the particle's path was actually observed by the detecting screen. This measurement is crucial for designing experiments that detect scattered particles and analyze their distribution.

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Most popular questions from this chapter

The differential cross section for scattering 6.5-MeV alpha particles at 120^ off a silver nucleus is about 0.5 barns/sr. If a total of \(10^{10}\) alphas impinge on a silver foil of thickness \(1 \mu \mathrm{m}\) and if we detect the scattered particles using a counter of area \(0.1 \mathrm{mm}^{2}\) at \(120^{\circ}\) and \(1 \mathrm{cm}\) from the target, about how many scattered alphas should we expect to count? (Silver has a specific gravity of \(10.5,\) and atomic mass of 108.)

(a) A certain nucleus has radius \(5 \mathrm{fm}\). ( \(1 \mathrm{fm}=10^{-15} \mathrm{m}\).) Find its cross section \(\sigma\) in barns. (1 barn \(=10^{-28} \mathrm{m}^{2} .\) ) (b) Do the same for an atom of radius \(0.1 \mathrm{nm}\). \(\left(1 \mathrm{nm}=10^{-9} \mathrm{m} .\right)\)

By integrating the element of solid angle \((14.15), d \Omega=\sin \theta d \theta d \phi\), over all directions, verify that the solid angle corresponding to all directions is 4 \(\pi\) steradians.

The cross section for scattering a certain nuclear particle by a nitrogen nucleus is 0.5 barns. If \(10^{11}\) of these particles are fired through a cloud chamber of length \(10 \mathrm{cm},\) containing nitrogen at STP, how many particles are scattered? (Use the ideal gas law and remember that each nitrogen molecule has two atoms. The scattering by any atomic electrons is completely negligible.)

A beam of particles is directed through a tank of liquid hydrogen. If the tank's length is \(50 \mathrm{cm}\) and the liquid density is 0.07 gram/cm \(^{3}\), what is the target density (number/area) of hydrogen atoms seen by the incident particles?

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