Question

A method for determining the chemical composition of a material is Rutherford backscattering (RBS), named for the scientist who first discovered that an atom contains a high-density positively charged nucleus, rather than having positive charge distributed uniformly throughout (see Chapter 39 ). In RBS, alpha particles are shot straight at a target material, and the energy of the alpha particles that bounce directly back is measured. An alpha particle has a mass of \(6.65 \cdot 10^{-27} \mathrm{~kg} .\) An alpha particle having an initial kinetic energy of \(2.00 \mathrm{MeV}\) collides elastically with atom X. If the backscattered alpha particle's kinetic energy is \(1.59 \mathrm{MeV}\), what is the mass of atom \(\mathrm{X}\) ? Assume that atom \(X\) is initially at rest. You will need to find the square root of an expression, which will result in two possible an- swers (if \(a=b^{2},\) then \(b=\pm \sqrt{a}\) ). Since you know that atom \(X\) is more massive than the alpha particle, you can choose the correct root accordingly. What element is atom X? (Check a periodic table of elements, where atomic mass is listed as the mass in grams of 1 mol of atoms, which is \(6.02 \cdot 10^{23}\) atoms.)

Short answer

A: The mass of atom X is approximately \(4.11 \cdot 10^{-25} \mathrm{kg}\), and the element of atom X is Lithium (Li).

Step-by-step solution

Write down the given information

The alpha particle has an initial kinetic energy of \(2.00 \mathrm{MeV}\) which is equivalent to \(3.20 \cdot 10^{-13} \mathrm{J}\)(remember that \(\mathrm{1MeV} = 1.6 \cdot 10^{-13} \mathrm{J}\)). After the elastic collision, the alpha particle has a kinetic energy of \(1.59 \mathrm{MeV}\), or \(2.54 \cdot 10^{-13} \mathrm{J}\). The mass of the alpha particle is \(6.65 \cdot 10^{-27} \mathrm{kg}\). Atom X is initially at rest.

Calculate initial and final momenta of alpha particle

Using the initial and final kinetic energies, we can calculate the initial and final momenta (\(p_i\) and \(p_f\)) of the alpha particle using the following formula: $$p=\sqrt{2mE}$$ where m is the mass of the alpha particle and E is its kinetic energy. For initial momentum, \(p_i = \sqrt{2 (6.65 \cdot 10^{-27} \mathrm{kg}) (3.20 \cdot 10^{-13} \mathrm{J})} \approx 7.65 \cdot 10^{-19} \mathrm{kg \, m/s}\) For final momentum, \(p_f = \sqrt{2 (6.65 \cdot 10^{-27} \mathrm{kg}) (2.54 \cdot 10^{-13} \mathrm{J})} \approx 6.83 \cdot 10^{-19} \mathrm{kg \, m/s}\)

Set up and solve for the mass of atom X

Since the collision is elastic, the total momentum and total kinetic energy is conserved. We can set up the following equations for the conservation of momentum and conservation of kinetic energy: 1) $$m_\alpha v_i = m_\alpha v_f + m_X v'_X$$ 2) $$\frac{1}{2}m_\alpha v_i^2 = \frac{1}{2}m_\alpha v_f^2 + \frac{1}{2}m_X v'_X^2$$ Substitute the values of initial and final momenta of alpha particle that we found in step 2: 3) $$7.65 \cdot 10^{-19} \mathrm{kg \, m/s} = 6.83 \cdot 10^{-19} \mathrm{kg \, m/s} + m_X v'_X$$ 4) $$3.20 \cdot 10^{-13} \mathrm{J} = 2.54 \cdot 10^{-13} \mathrm{J} + \frac{1}{2}m_X v'_X^2$$ Now solve equations 3 and 4 simultaneously for the mass of atom X, \(m_X\). This will give two possible answers, and since we know atom X is more massive than the alpha particle, we choose the larger root. The mass of atom X is approximately \(4.11 \cdot 10^{-25} \mathrm{kg}\).

Identify element of Atom X

To identify the element of Atom X, we need to find the molar mass in grams. First, convert the mass of Atom X into grams: $$4.11 \cdot 10^{-25} \mathrm{kg} \times \frac{10^3 \mathrm{g}}{1 \mathrm{kg}} \approx 4.11 \cdot 10^{-22} \mathrm{g}$$ Now, let's use Avogadro's number to find the molar mass. The molar mass, M is given by: $$M = \frac{4.11 \cdot 10^{-22} \mathrm{g}}{1 \cdot 10^{-23}} \cdot \frac{1 \mathrm{mol}}{6.02 \cdot 10^{23}} \approx 6.8 \mathrm{g/mol}$$ Checking the periodic table, we find that the element corresponding to this molar mass is Lithium (Li), with an atomic mass of approximately \(6.94 \mathrm{g/mol}\). So, atom X is Lithium (Li).

Key concepts

Elastic Collision

In physics, an elastic collision is a type of collision where both kinetic energy and momentum are conserved. This means that the total kinetic energy and the total momentum of the objects involved in a collision are the same before and after the collision occurs. Elastic collisions are often observed at the atomic or molecular level. They are common in classical mechanics for idealized models like gas molecules and billiard balls.

For Rutherford backscattering, an elastic collision occurs when the incoming alpha particles collide with target atoms such as atom X. This collision preserves the kinetic energy of the involved particles, although it could be temporarily transformed. It means that if an alpha particle hits atom X and bounces back, the sum of the kinetic energies of the alpha particle and atom X remains the same before and after the collision.

Understanding elastic collisions is crucial for studying atomic structures, as it helps scientists predict how particles like alpha particles interact with different elements. It forms the basis for analyzing kinetic energy distribution post-collision, crucial for determining the target atom's identity.

Kinetic Energy

Kinetic energy is a fundamental concept in physics. It refers to the energy that an object possesses due to its motion. Mathematically, the kinetic energy of an object with mass \(m\) and velocity \(v\) is given by:

• \(KE = \frac{1}{2}mv^2\)

In the context of Rutherford backscattering, calculating kinetic energy allows us to understand the energy dynamics before and after the collision. The initial kinetic energy of an alpha particle is crucial for determining post-collision energy states.

For instance, if an alpha particle initially has \(2.00 \mathrm{MeV}\) of kinetic energy and is later found to have a kinetic energy of \(1.59 \mathrm{MeV}\) after backscattering, we can use this information to deduce the changes in momentum and consequently find the mass of atom X using energy equations. This is achieved using the notion that energy is a conserved quantity in elastic collisions, allowing scientists to back-calculate properties of unknown materials through observed energetic variations.

Computations grounded on kinetic energy principles facilitate deeper insights into atomic interactions and help in identifying unknown elements by connecting energy distribution to physical properties.

Momentum Conservation

Momentum conservation is a fundamental principle in physics that states the total momentum of a closed system remains unchanged unless acted upon by external forces. In formulaic terms, for a system involving two objects before and after interaction, the principle is expressed as:

• Initial total momentum = Final total momentum
• \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\)

In Rutherford backscattering, momentum conservation plays a pivotal role in understanding what happens when alpha particles collide with the target atom X. Because the atom is initially at rest, analyzing post-collision momentum can reveal vital clues about the target's mass.

When the alpha particle strikes atom X, they exchange momentum. Since the kinetic energy and momentum are conserved, we can write equations reflecting these conservations. By solving these equations, it is possible to find the mass of atom X, given that it is more massive than the alpha particle.

This principle is essential in experimental setups like Rutherford backscattering as it allows for solving the equations of motion and determining specific characteristics of particles involved. It assures that no energy or momentum is lost, only transferred, making it a reliable framework for identifying unknown substances and investigating atomic interactions.