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Q56P

Expert-verifiedFound in: Page 1305

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An electron is emitted from a middle-mass nuclide (A=150, say) with a kinetic energy of 1.0 MeV. (a) What is its de-Broglie wavelength? (b) Calculate the radius of the emitting nucleus. (c) Can such an electron be confined as a standing wave in a “box” of such dimensions? (d) Can you use these numbers to disprove the (abandoned) argument that electrons actually exist in nuclei?**

- The de-Broglie wavelength of the electron is $9\times {10}^{2}\mathrm{fm}$ .
- The radius of the emitting nucleus is 6.4 fm .
- No, the electron cannot be confined as a standing wave in a “box” of such dimensions due to their large wavelengths.
- Yes, these numbers can be used to disprove the argument that electrons actually exist.

- Mass number of the nuclide A = 150.
- Kinetic energy of the emitted electron.

**The electron can only be confined within a given structure only if its wavelength is smaller than the atomic radius of the nuclide. Now, the wavelength of an electron is associated with the momentum according to the de-Broglie concept. Thus, by comparing the wavelength with the calculated atomic radius of the nuclide, consider the strong argument for the case of confinement.**

The kinetic energy of a particle in motion:

$K=m{c}^{2}$ ….. (i)

The energy and momentum relation according to relativistic concept,

$pc=\sqrt{{K}^{2}+{m}^{2}{c}^{4}}$ …… (ii)

The atomic radius of a nuclide using its nucleon or mass number,

$r={r}_{0}{A}^{\frac{1}{3}}$ …… (iii)

The de-Broglie wavelength of a particle of smaller size:

$\lambda =\frac{h}{p}$ …… (iv)

Consider the known data, ${\mathrm{mc}}^{2}=0.511\mathrm{MeV}$ and hc = 1240 MeV.fm

Substitute the value of momentum from equation (ii) in equation (iv), consider de-Broglie wavelength of the electron as follows:

$\lambda =\frac{hc}{\sqrt{{K}^{2}+{m}^{2}{c}^{4}}}\phantom{\rule{0ex}{0ex}}=\frac{hc}{\sqrt{{K}^{2}+2Km{c}^{2}}}\left(\mathrm{from}\mathrm{equation}(\mathrm{i}\right),\mathrm{K}={\mathrm{mc}}^{2})$

Substitute the values as:

$\mathrm{\lambda}=\frac{1240\mathrm{MeV}.\mathrm{fm}}{\sqrt{{\left(1.0\mathrm{MeV}\right)}^{2}+2\left(1.0\mathrm{MeV}\right)\left(0.511\mathrm{MeV}\right)}}\phantom{\rule{0ex}{0ex}}=9\times {10}^{2}\mathrm{fm}$

Hence, the value of the wavelength is $9\times {10}^{2}\mathrm{fm}$ .

Using the given data in equation (iii), determine the radius of the emitting nucleus as follows:

$\mathrm{r}=\left(1.2\mathrm{fm}\right){\left(150\right)}^{1/3}\phantom{\rule{0ex}{0ex}}=6.4\mathrm{fm}$

Hence, the value of the radius is 6.4 fm .

Since, $\lambda >>r$ from parts (a) and (b) calculations* *the electron cannot be confined in the nuclide. Recall that at least $\frac{\lambda}{2}$ is needed in any particular direction, to support a standing wave in an “infinite well.” A finite well is able to support *slightly *less than $\frac{\lambda}{2}$ (as one can infer from the ground state wave function in Fig. 39-6), but in the present case* $\frac{\lambda}{r}$ *is far too big to be supported.

A strong cas e can be made on the basis of the remarks in part (c), above.

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