Problem 75 - Chemistry: The Central Science

Question

The nuclear masses of \({ }^{7} \mathrm{Be},{ }^{9} \mathrm{Be}\), and \({ }^{10} \mathrm{Be}\) are \(7.0147\), \(9.0100\), and \(10.0113\) amu, respectively. Which of these nuclei has the largest binding energy per nucleon?

Step-by-Step Solution

Answer

The isotope with the largest binding energy per nucleon is \({ }^{10}\mathrm{Be}\) with a binding energy per nucleon of 1.043724 x 10^-12 J/nucleon, making it the most stable nucleus among the given isotopes of Be.

Step 1: Calculate the mass defect for each isotope

To find the mass defect, subtract the actual mass of the nucleus from the sum of the masses of its constituent protons and neutrons. The masses of a proton and neutron are approximately 1.007276 amu and 1.008665 amu, respectively. For \({ }^{7}\mathrm{Be}\), there are 4 protons and 3 neutrons: Mass defect = (4 * 1.007276) + (3 * 1.008665) - 7.0147 Mass defect = 4.029104 + 3.025995 - 7.0147 Mass defect = 7.055099 - 7.0147 = 0.040399 amu For \({ }^{9}\mathrm{Be}\), there are 4 protons and 5 neutrons: Mass defect = (4 * 1.007276) + (5 * 1.008665) - 9.0100 Mass defect = 4.029104 + 5.043325 - 9.0100 Mass defect = 9.072429 - 9.0100 = 0.062429 amu For \({ }^{10}\mathrm{Be}\), there are 4 protons and 6 neutrons: Mass defect = (4 * 1.007276) + (6 * 1.008665) - 10.0113 Mass defect = 4.029104 + 6.051990 - 10.0113 Mass defect = 10.081094 - 10.0113 = 0.069794 amu

Step 2: Calculate the binding energy for each isotope

Using the mass defect values calculated in Step 1, we can now determine the binding energy by converting the mass defect to energy using the Einstein's equation, E=mc^2, where the mass m is in kilograms and the speed of light c is 3.00 x 10^8 m/s. First, we need to convert the mass defect from amu (atomic mass units) to kilograms using the conversion factor 1 amu = 1.66054 x 10^-27 kg. For \({ }^{7}\mathrm{Be}\): Mass defect in kg = 0.040399 amu * 1.66054 x 10^-27 kg/amu Mass defect in kg = 6.711949 x 10^-29 kg Binding energy = E = mc^2 Binding energy = (6.711949 x 10^-29 kg)(3.00 x 10^8 m/s)^2 Binding energy = 6.040162 x 10^-12 J For \({ }^{9}\mathrm{Be}\): Mass defect in kg = 0.062429 amu * 1.66054 x 10^-27 kg/amu Mass defect in kg = 1.036668 x 10^-28 kg Binding energy = E = mc^2 Binding energy = (1.036668 x 10^-28 kg)(3.00 x 10^8 m/s)^2 Binding energy = 9.330076 x 10^-12 J For \({ }^{10}\mathrm{Be}\): Mass defect in kg = 0.069794 amu * 1.66054 x 10^-27 kg/amu Mass defect in kg = 1.159459 x 10^-28 kg Binding energy = E = mc^2 Binding energy = (1.159459 x 10^-28 kg)(3.00 x 10^8 m/s)^2 Binding energy = 1.043724 x 10^-11 J

Step 3: Calculate the binding energy per nucleon for each isotope

Now divide the binding energy by the number of nucleons (protons and neutrons) in each isotope to get the binding energy per nucleon. For \({ }^{7}\mathrm{Be}\): Binding energy per nucleon = (6.040162 x 10^-12 J) / 7 nucleons Binding energy per nucleon = 8.628803 x 10^-13 J/nucleon For \({ }^{9}\mathrm{Be}\): Binding energy per nucleon = (9.330076 x 10^-12 J) / 9 nucleons Binding energy per nucleon = 1.036675 x 10^-12 J/nucleon For \({ }^{10}\mathrm{Be}\): Binding energy per nucleon = (1.043724 x 10^-11 J) / 10 nucleons Binding energy per nucleon = 1.043724 x 10^-12 J/nucleon

Step 4: Determine which isotope has the largest binding energy per nucleon

Compare the binding energy per nucleon values calculated in Step 3. The isotope with the largest binding energy per nucleon is the most stable isotope among the given ones. \({ }^{7}\mathrm{Be}\): 8.628803 x 10^-13 J/nucleon \({ }^{9}\mathrm{Be}\): 1.036675 x 10^-12 J/nucleon \({ }^{10}\mathrm{Be}\): 1.043724 x 10^-12 J/nucleon The largest binding energy per nucleon is for \({ }^{10}\mathrm{Be}\), which indicates that it is the most stable nucleus among the given isotopes of Be.

Key Concepts

Mass Defect

Mass defect is a fascinating concept in nuclear physics that explains the difference in mass between the sum of the individual protons and neutrons in a nucleus and the actual mass of the nucleus itself. Protons and neutrons are both quite hefty in terms of mass, with the mass of a proton being approximately 1.007276 atomic mass units (amu), and a neutron weighing in at about 1.008665 amu.

When you calculate the total mass of the protons and neutrons separately and then subtract the measured atomic mass, you get the mass defect. Let's look at the element beryllium (Be) as an example. For its isotopes

• extsuperscript{7} ext{Be}, you would expect a sum of masses for the 4 protons and 3 neutrons to be more than its actual nuclear mass; hence the mass defect.
• Similarly, for extsuperscript{9} ext{Be}, there’s a notable discrepancy when you compare its combined mass of 4 protons and 5 neutrons.
• extsuperscript{10} ext{Be is also interesting as it features a mass defect from its 6 neutrons in addition to the 4 protons.

The mass defect is key to understanding atomic stability, influencing the concept of binding energy, which involves converting this mass deficit into energy through Einstein's famous equation.

Atomic Mass Unit (amu)

An atomic mass unit (amu) is a standard unit of mass that quantifies the amount of mass in an atom or particle. It’s a small unit, reflecting the tiny weights of atoms. Essentially, it helps make calculations in atomic physics manageable.

• 1 amu is presently defined as one twelfth of the mass of a carbon-12 atom.
• This unit is crucial when dealing with protons and neutrons since each of these subatomic particles weighs about 1 amu.
• It allows scientists to precisely measure and communicate atomic masses, facilitating understanding when calculating mass defects and binding energy.

Using the mass unit makes it more straightforward to handle the minuscule mass defects and turn them into more tangible concepts like energy. While the mass of everyday objects is usually expressed in grams or kilograms, the atomic mass unit is the go-to in atomic-level measurements, emphasizing its indispensability in nuclear studies.

Einstein's Equation (E=mc^2)

Einstein's iconic equation, which states that energy (E) is equal to mass (m) multiplied by the speed of light (c) squared, provides a pivotal framework for understanding the mass defect. Through this equation: \[ E = mc^2 \] we understand that the mass defect in nuclei is actually converted and released as energy, accounting for atomic stability.

• The "c" in this equation represents the speed of light, a whopping 3.00 x 10^8 m/s; indicating how a small amount of mass can generate a vast amount of energy.
• This fundamental principle shows the relationship between mass loss and energy released during nuclear processes.
• Einstein's equation is not just a theoretical notion but a practical tool used in nuclear physics to calculate binding energy.

By applying this equation, the mass defects previously calculated in atomic mass units are converted into the binding energy, with results usually given in Joules. This profound relationship outlined by Einstein has had far-reaching impacts beyond physics, shaping the way we view energy production and atomic interactions.