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?
The energy of the given electrons, K = 200 MeV
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.
The de-Broglie wavelength of a body,
Where, p is the momentum and h is Planck’s constant.
The relativistic relation between the energy and momentum,
Here, m is the mass and c is the speed of light.
The kinetic energy of a body in motion with speed of light,
With the given data, equation (iii) and in equation (ii), you can get the value as follows:
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:
Hence, the value of wavelength is 6.2 fm.
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.
(a) Show that the total binding energy of a given nuclide is , where, is the mass excess of , is the mass excess of a neutron, and is the mass excess of the given nuclide. (b) Using this method, calculate the binding energy per nucleon for . Compare your result with the value listed in Table 42-1. The needed mass excesses, rounded to three significant figures, are , , and . Note the economy of calculation that results when mass excesses are used in place of the actual masses.
Because of the 1986 explosion and fire in a reactor at the Chernobyl nuclear power plant in northern Ukraine, part of Ukraine is contaminated with ,which undergoes beta-minus decay with a half-life of 30.2y . In 1996, the total activity of this contamination over an area of was estimated to be . Assume that the is uniformly spread over that area and that the beta-decay electrons travel either directly upward or directly downward. How many beta-decay electrons would you intercept were you to lie on the ground in that area for (a) in 1996 and (b) today? (You need to estimate your cross-sectional area that intercepts those electrons.)
a. Show that the mass M of an atom is given approximately by , where A is the mass number and is the proton mass. For (b) , (c),(d) , (e) , and (f) , use Table 42-1 to find the percentage deviation between and :
(g) Is a value of accurate enough to be used in a calculation of a nuclear binding energy?
A certain radionuclide is being manufactured in a cyclotron at a constant rate R. It is also decaying with disintegration constant . Assume that the production process has been going on for a time that is much longer than the half-life of the radionuclide. (a) Show that the numbers of radioactive nuclei present after such time remains constant and is given by. (b) Now show that this result holds no matter how many radioactive nuclei were present initially. The nuclide is said to be in secular equilibrium with its source; in this state its decay rate is just equal to its production rate.
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