Question

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?

Short answer

1. The de-Broglie wavelength of the electron is$9\times {10}^2\mathrm{fm}$ .
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.

Step-by-step solution

The given data

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

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)

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:

$$\begin{gathered} \lambda =\frac{hc}{\sqrt{K^2+m^2{c}^4}} \\ =\frac{hc}{\sqrt{K^2+2Km{c}^2}}\left(\mathrm{from}\mathrm{equation}(\mathrm{i}\right),\mathrm{K}={\mathrm{mc}}^2) \end{gathered}$$

Substitute the values as:

$$\begin{gathered} \mathrm{\lambda }=\frac{1240\mathrm{MeV}.\mathrm{fm}}{\sqrt{{\left(1.0\mathrm{MeV}\right)}^2+2\left(1.0\mathrm{MeV}\right)\left(0.511\mathrm{MeV}\right)}} \\ =9\times {10}^2\mathrm{fm} \end{gathered}$$

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

b) Calculate the radius of the emitting nucleus

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

$$\begin{gathered} \mathrm{r}=\left(1.2\mathrm{fm}\right){\left(150\right)}^{1/3} \\ =6.4\mathrm{fm} \end{gathered}$$

Hence, the value of the radius is 6.4 fm .

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

Since,$\lambda >>r$from parts (a) and (b) calculationsthe 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 supportslightlyless than $\frac{\lambda }{2}$(as one caninfer 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.

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.