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A beam of particles is directed through a tank of liquid hydrogen. If the tank's length is 50cm 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?

Short Answer

Expert verified
The target density is approximately 2.09×1024atoms/cm2.

Step by step solution

01

Calculate the volume of the hydrogen tank

Since the tank's length is given as 50cm, the volume of hydrogen through which the particles pass is essentially a column of liquid with a length of 50cm. We consider the cross-sectional area of the tank to be 1cm2 for simplicity, so the volume V is 50cm3.
02

Calculate the mass of hydrogen in the tank

Given the density ρ is 0.07g/cm3 and the volume V is 50cm3, the mass m of hydrogen is calculated by m=ρ×V=0.07g/cm3×50cm3=3.5g.
03

Determine the number of hydrogen molecules

The number of moles n of hydrogen is 3.5g2.016g/mol, because the molecular mass of H2 is approximately 2.016g/mol. Hence, n1.7357mol.
04

Calculate the number of hydrogen atoms

Since one molecule of H2 contains two hydrogen atoms, multiply the number of molecules by Avogadro's number (6.022×1023molecules/mol). The total number of hydrogen atoms N is n×2×6.022×1023. Thus, N2.09×1024atoms.
05

Calculate the target density (number/area)

The target density is the number of atoms per unit area. Given the cross-sectional area is 1cm2 for which these calculations were simplified, the target density is simply the total number of atoms N, which is 2.09×1024atoms/cm2.

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Key Concepts

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

Understanding Target Density
In particle physics, calculating the target density is crucial for understanding how particles interact with a medium when they pass through it. Target density refers to the number of atoms or molecules that are present within a certain area through which a beam of particles travels. This concept helps researchers predict collision events and understand interactions at a microscopic level.

The calculation in the exercise simplifies this by assuming we have a basic geometric setup—a column with a known cross-sectional area and length. We define target density mathematically as the number of atoms per unit area. It’s calculated by determining the total number of atoms (or particles) in the column and dividing this by the cross-sectional area. This allows physicists to estimate how many particles in the beam are likely to interact with the atoms in the tank. This measure is vital for experimental arrangements in fields like nuclear medicine, astrophysics, and accelerator physics.
Exploring Hydrogen Atoms
Hydrogen atoms, the simplest and most abundant elements in the universe, consist of just one proton and one electron. In particle physics, they play a significant role due to their simplicity and prevalence.

When we calculate target density, understanding the structure of hydrogen is pivotal. In the exercise, a tank filled with liquid hydrogen is utilized. Here, hydrogen exists in the molecular form, which means two atoms form a H2 molecule. Calculations must consider that each mole of H2 consists of two hydrogen atoms. This impacts the step where we use the molecular weight to determine the number of moles in the hydrogen mass.
  • The mass of one mole of H2 is 2.016 g/mol.
  • Avogadro’s number, 6.022×1023molecules/mol, is used to determine the number of molecules, knocking twice for atoms.
Such transitions from molecules to atoms are central in simulations and practical experiments involving hydrogen, including studies on energy production through fusion.
Liquid Density in Particle Physics
Liquid density is a foundational concept that frequently appears in calculations concerning particle interactions in liquids. It denotes the mass of a substance per unit volume and signifies how closely packed those particles, like atoms or molecules, are.

In the context of the exercise, liquid hydrogen is the medium through which particles travel. The density is given as 0.07g/cm3, helping to calculate the overall mass of hydrogen in the tank. This density affects how particles in the beam scatter when passing through, impacting the experimental results and models used for analysis.
  • It determines how resistive the medium is to particle movement.
  • Helps in calculating the number of interacting particles or atoms.
Such considerations are crucial when setting up conditions for experiments, ensuring that the test results are accurate and reflective of the underlying physics principles. Liquid density impacts turbulence and reactions within the medium, and therefore, forms a bridge to understanding complex systems in particle physics.

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Most popular questions from this chapter

Calculate the solid angles subtended by the moon and by the sun, both as seen from the earth. Comment on your answers. (The radii of the moon and sun are Rm=1.74×106m and Rs=6.96×108 m. Their distances from earth are dm=3.84×108m and ds=1.50×1011m.)

Our definition of the scattering cross section, Nsc=Ninc ntar σ, applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case Nsc=ninc Ntarσ where ninc  is the density (number/area) of the incident beam, viewed head-on, and Ntar  is the total number of targets in the target assembly.

By integrating the element of solid angle (14.15),dΩ=sinθdθdϕ, over all directions, verify that the solid angle corresponding to all directions is 4 π steradians.

(a) A certain nucleus has radius 5fm. ( 1fm=1015m.) Find its cross section σ in barns. (1 barn =1028m2. ) (b) Do the same for an atom of radius 0.1nm. (1nm=109m.)

A blueberry pancake has diameter 15 cm and contains 6 large blueberries, each of diameter 1 cm. Find the cross section σ of a blueberry and the "target" density ntar  (number/area) of berries in the pancake, as seen from above. What is the probability that a skewer, jabbed at random into the pancake, will hit a berry (in terms of σ and ntar  and then numerically)?

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