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Nuclear and Particle Physics Assignment

1) The mass of 1mm3 of nuclear matter is vastly greater than the mass of the Earth. 2) The ratio of the Coulomb to gravitational force between two protons decreases rapidly with increasing distance between them. 3) For a uniformly charged sphere, the mean square charge radius is equal to 3/5 the square of the sphere's actual radius.

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344 views1 page

Nuclear and Particle Physics Assignment

1) The mass of 1mm3 of nuclear matter is vastly greater than the mass of the Earth. 2) The ratio of the Coulomb to gravitational force between two protons decreases rapidly with increasing distance between them. 3) For a uniformly charged sphere, the mean square charge radius is equal to 3/5 the square of the sphere's actual radius.

Uploaded by

Vinayak Raj
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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PHY-526 A (Introduction to Nuclear and Particle Physics)

Practice Assignment - 1 (Date: - 09/01/2017)

1. Compare the mass of 1mm3 of nuclear matter and the mass of Earth.

2. Calculate the ratio of the Coulomb and gravitational forces between two
protons. How does this ratio varies with distance between them?

3. Show that the mean square charge radius of a uniformly charged sphere is
3
< 𝑟2 > = 5
𝑅2

4. Estimate the de-Brogli wavelength of a 420 MeV electron.

5. Using uncertainty principle, estimate the minimum kinetic energy of an


electron confined within a nucleus of size 10 fm.

6. If the kinetic energy of an neutron, confined inside a cubic box, is 10 MeV in


its ground state, calculate the size of the box. Also calculate the energy of the
first excited state.

7. Find the Coulomb energy of a uniformly charged sphere. Assuming nucleus to


be a uniformly charged sphere, estimate the electrostatic energy difference
1
11 11
between nuclei 5𝐵 and 6𝐶 . You may take 𝑅𝑁 = 𝑅0 𝐴3 and 𝑅0 = 1.2 𝑓𝑚.

8. a) Show that the interaction potential between an electron and a uniformly


charged sphere is given by
𝑍𝑒 2 3 1 𝑟 2
𝑉 ′ (𝑟) = − 4𝜋𝜀 �2 − � � � for r < R, where R is the nuclear radius.
0𝑅 2 𝑅
b) The actual interaction potential for an atomic electron due to a uniformly
charged sphere can be written as 𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑉 ′ (𝑟 < 𝑅) + 𝑉(𝑟 > 𝑅), where V’ is
as above and V is the usual coulomb potential due to a point nucleus. If the
effect of the finite size of a nucleus is only to change the energy of the
electronic states relative to the point nucleus value by< ∆𝐸 > = < 𝑉𝑡𝑜𝑡𝑎𝑙 >
− < 𝑉 >, using the 1S hydrogenic wave function, show that

2 𝑍4 𝑒 2𝑅2
∆𝐸 = � �
5 4𝜋𝜀0 𝑎03

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