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### Fundamentals Of Physics

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?

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

## Step 1: The given data

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

## Step 2: Understanding the concept of size and confinement

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)

## Step 3: a) Calculate the de-Broglie wavelength

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:

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

Substitute the values as:

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

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

## Step 4: b) Calculate the radius of the emitting nucleus

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 .

## Step 5: c) Calculate whether the electron of the given nuclide can be confined as a standing wave

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.

## Step 6: d) Determine whether an argument can be raised for part (c)

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