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Q51P

Expert-verifiedFound in: Page 1304

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A**** ${}^{{\mathbf{238}}}{\mathbf{\cup}}$ nucleus emits an**** 4.196 MeV alpha particle. Calculate the disintegration energy*** Q ***for this process, taking the recoil energy of the residual ${}^{{\mathbf{238}}}{\mathit{T}}{\mathit{h}}$**** nucleus into account.**

The disintegration energy for the process of a uranium nucleus emitting an alpha particle is 4.269 MeV.

Energy of the alpha particle, ${E}_{\alpha}=4.196MeV$

**The disintegration energy, also known as the Q-value, is the energy that is absorbed or released when a nuclear reaction takes place. The recoiling energy of the thorium is the translational energy of the thorium nucleus. Considering the energy and momentum conservation concept, we can get the value of this recoiling energy. Again, the total energy released by the alpha particle and thorium will give the disintegration energy.**

**Formula:**

**The kinetic energy of the particle according to classical concept, ${\mathit{K}}{\mathbf{=}}\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$**** **** (i)**

Energy and momentum are conserved. We assume the residual thorium nucleus is in its ground state.

Let, ${K}_{\alpha}$ be the kinetic energy of the alpha particle and ${K}_{Th}$ be the kinetic energy of the thorium nucleus. Then, the disintegration energy for the process emitting the alpha particle and thorium nuclide is given as:

$Q={K}_{\alpha}+{K}_{Th}...........................\left(a\right)$

We assume the uranium nucleus is initially at rest. Then, conservation of momentum yields

$\begin{array}{l}0={p}_{\alpha}+{p}_{Th}\\ {p}_{Th}=-{p}_{\alpha}\dots \dots \dots ....\dots \dots \dots \left(b\right)\end{array}$

Where, ${p}_{\alpha}$ is the momentum of the alpha particle and ${p}_{Th}$ is the momentum of the thorium nucleus.

Both particles travel slowly enough that the classical relationship between momentum and energy can be used.

Thus, using equation (i), the kinetic energy of the thorium nucleus is given as:

${K}_{Th}=\frac{{p}_{Th}}{2{m}_{Th}}$

Now, using the value of momentum from equation (a) and using the kinetic energy value from equation (i), we get the above value as:

${K}_{Th}=\frac{{P}_{\alpha}^{2}}{2{m}_{Th}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{\alpha}}{{m}_{Th}}{K}_{\alpha}$

Now, using the above value in equation (a), we can get the disintegration energy as follows:

$Q={K}_{\alpha}+\frac{{m}_{\alpha}}{{m}_{Th}}{K}_{\alpha}\phantom{\rule{0ex}{0ex}}=4.196MeV+\left(\frac{4u}{234u}\right)\left(4.196MeV\right)\phantom{\rule{0ex}{0ex}}=4.196MeV+0.07313MeV\phantom{\rule{0ex}{0ex}}\approx 4.269MeV$

Hence, the value of the disintegration energy is $4.269MeV$.

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