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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Nuclear radii may be measured by scattering high energy (high speed) electrons from nuclei. (a) What is the de-Broglie wavelength for 200MeV electrons? (b) Are these electrons suitable probes for this purpose?

1. The de-Broglie wavelength for electrons is 6.2 fm.
2. Yes, these electrons are suitable for the purpose of lower energy probes.
See the step by step solution

## Step 1: The given data:

The energy of the given electrons, K = 200 MeV

## Step 2: Understanding the concept of the relativistic case:

You can get the detail of the nucleus if only the de-Broglie wavelength is smaller than that compared to the size or the radius of the nucleus. Since the kinetic energy K of the electron is much greater than its rest energy, a relativistic formulation must be used. Thus, using the relativistic approach to the concept of kinetic energy and momentum relation, we can get the required relativistic energy value that will further help in getting the de-Broglie wavelength.

Formulae:

The de-Broglie wavelength of a body,

$\lambda =\frac{h}{p}$ ….. (i)

Where, p is the momentum and h is Planck’s constant.

The relativistic relation between the energy and momentum,

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

Here, m is the mass and c is the speed of light.

The kinetic energy of a body in motion with speed of light,

$K=\frac{1}{2}m{c}^{2}$ ….. (iii)

## Step 3: (a) Calculation of the de-Broglie wavelength:

With the given data, equation (iii) and ${\mathrm{mc}}^{2}=0.511\mathrm{MeV}$ in equation (ii), you can get the value as follows:

$\mathrm{pc}=\sqrt{{\mathrm{K}}^{2}+2{\mathrm{Kmc}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(200\mathrm{MeV}\right)}^{2}+2\left(200\mathrm{MeV}\right)\left(0.511\mathrm{MeV}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{40000{\mathrm{MeV}}^{2}+204.4{\mathrm{MeV}}^{2}}\phantom{\rule{0ex}{0ex}}=200.5\mathrm{MeV}$

Thus, using the above value in equation (i) after multiplying in both the numerator and denominator, we can get the de-Broglie wavelength value as follows:

$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{pc}}\phantom{\rule{0ex}{0ex}}=\frac{1240\mathrm{eV}.\mathrm{nm}}{200.5×{10}^{6}\mathrm{eV}}\phantom{\rule{0ex}{0ex}}=6.18×{10}^{-6}\mathrm{nm}\phantom{\rule{0ex}{0ex}}\approx 6.2\mathrm{fm}$

Hence, the value of wavelength is 6.2 fm.

## Step 4: (b) Calculation to know about the suitable electrons for the probe:

The diameter of a copper nucleus, for example, is about 8.6fm , just a little larger than the de Broglie wavelength of an 200 MeV electron. To resolve detail, the wavelength should be smaller than the target, ideally a tenth of the diameter or less. The 200 MeV electrons are perhaps at the lower limit in energy for useful probes.

The more energetic the incident particle; the finer are the details of the target that can be probed.

Hence, these electrons are suitable for the purpose of lower energy probes.

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