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.)

Step by step solution

01

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.

02

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, \(p_f = \sqrt{2 (6.65 \cdot 10^{-27} \mathrm{kg}) (2.54 \cdot 10^{-13} \mathrm{J})} \approx 6.83 \cdot 10^{-19} \mathrm{kg \, m/s}\)

03

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 \(4.11 \cdot 10^{-25} \mathrm{kg}\).

04

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).

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