Question

Calculate \(\Delta \mathrm{m}\), the difference in mass between the final and initial nuclei in grams per mole for the emission of a \(\gamma\) ray, \({ }^{19}{ }_{8} \mathrm{O}^{*}(\) excited state \() \rightarrow{ }^{19} \mathrm{O} \mathrm{O}\) (ground state \()+\gamma\) \(\left(\Delta \mathrm{E}=1.06 \times 10^{8} \mathrm{Kcal} / \mathrm{mole}\right)\)

Short answer

The difference in mass between the final and initial nuclei for the emission of a gamma ray is approximately \(4.926\times10^{-3}\ \text{g/mole}\).

Step-by-step solution

Convert the energy change to joules per mole

The change in energy is given in kilocalories per mole, but we need to convert it to joules per mole to use Einstein's formula. We can do this using the conversion factor 1 kcal = 4184 J: \(\Delta E = (1.06 \times 10^8\ \text{kcal/mole}) \times (4184\ \text{J/kcal}) = 4.434\times10^{11}\ \text{J/mole}\)

Apply Einstein's mass-energy equivalence formula

Now, we can use the mass-energy equivalence formula to find the change in mass: \(\Delta E = \Delta m \times c^2\) We need to isolate \(\Delta m\), so divide both sides by \(c^2\): \(\Delta m = \frac{\Delta E}{c^2}\) The speed of light, \(c\), is approximately \(3 \times 10^8\ \text{m/s}\). Plug in the given values: \(\Delta m = \frac{4.434\times10^{11}\ \text{J/mole}}{(3\times10^{8}\ \text{m/s})^2}\)

Calculate the change in mass

Now, calculate the change in mass: \(\Delta m = \frac{4.434\times10^{11}\ \text{J/mole}}{9\times10^{16}\ \text{m}^2\text{/s}^2} = 4.926\times10^{-6}\ \text{kg/mole}\)

Convert the change in mass to grams per mole

Finally, convert the change in mass from kilograms per mole to grams per mole using the conversion factor 1 kg = 1000 g: \(\Delta m = (4.926\times10^{-6}\ \text{kg/mole}) \times (1000\ \text{g/kg}) = 4.926\times10^{-3}\ \text{g/mole}\) Thus, the difference in mass between the final and initial nuclei for the emission of a gamma ray is approximately \(4.926\times10^{-3}\ \text{g/mole}\).

Key concepts

Energy Conversion

Energy conversion is essential for understanding reactions and processes in physics, as it involves changing one form of energy into another. In this exercise, energy is initially given in kilocalories (kcal) but needs to be converted into joules (J) since the calculations involve physical constants measured in joules.

• Kilocalories are a unit of energy commonly used in chemistry and biology to quantify large energy changes, such as those involved in metabolic processes.
• Joules, on the other hand, are used in physics because they are the standard International System of Units (SI) for energy, making them useful for calculations involving fundamental physics constants.

In this problem, the energy conversion uses the factor 1 kcal = 4184 J. This conversion is crucial as it allows the use of Einstein's mass-energy equivalence formula, which is expressed in joules.

Gamma Ray Emission

Gamma rays are a form of electromagnetic radiation that result from nuclear transitions, meaning they originate within the atomic nucleus rather than involving electronic transitions like X-rays. In this problem, an excited nucleus emits a gamma ray to transition to a lower energy state.

• Gamma rays are high-energy photons and possess no mass or charge but have penetrating power.
• They are often emitted during decay processes in which a nucleus transitions from an excited to a more stable state, releasing energy in the form of gamma radiation.

In the emission event described in the exercise, a nucleus of oxygen ( ^{19}_{8}O^{*} ) sheds energy by emitting a gamma ray and settling into a ground state. This energy release is quantifiable, as seen through changes in mass due to the equivalence principle.

Nuclear Reactions

Nuclear reactions involve changes in an atom's nucleus and can result in the transformation of elements or isotopes, as well as the emission of energy. Such reactions are distinct from chemical reactions, which only involve electron interactions.

• Within the nucleus, altering the configuration of protons and neutrons usually requires or releases significant amounts of energy.
• In the given problem, the reaction involves a nuclear transition from an excited state to a ground state, illustrating a common type of nuclear reaction.

This energy release or absorption is central to nuclear physics and is described by the equation \(E=mc^2\), which relates energy changes to mass changes during nuclear phenomena.

Einstein's Formula

Einstein's formula, known as the mass-energy equivalence principle, is famously represented as \(E=mc^2\). This signifies that mass can be converted into energy and vice versa; a small change in mass results in a huge amount of energy due to the squared speed of light.

• The formula underlies the concept of why massive energy can be derived from nuclear and particle physics, evidenced by nuclear reactions and phenomena such as fusion and fission.
• In this exercise, we use the rearranged form \(\Delta m = \frac{\Delta E}{c^2}\) to find the tiny mass difference resulting from gamma-ray emission.

Understanding \(E=mc^2\) is crucial in high-energy physics, astrophysics, and applications such as nuclear power, highlighting the profound connection between mass and energy.