Question

In: Physics

The nuclide 236U decays with emission of an alpha particle with measured energy of 4.448 MeV...

The nuclide 236U decays with emission of an alpha particle with measured energy of 4.448 MeV and an accompanying gamma photon with measured energy of 49 keV. What is the product nucleus and does this particular decay of 236U occur to the ground state or and excited state of the product nuclide?

Expert Solution

a) The general equation of an alpha decay from a parent nucleus P to a daughter nucleus D is given by

Here, P = 236 -U

So, A+4 = 236,

A = 236-4 = 232

For Uranium, atomic number = 92

So, Z+2 = 92

Z = 90

The product will be

The atom with Z = 90 is Thorium.

So, the product is

b) Here, we can see that a gamma decay occurs, following the reaction.

The nuclei, that are formed after an alpha or beta decay are usually not in the ground state of the atom, but in some excited state.

The atom has to come back to its ground state by releasing some energy. This is done by a gamma decay.

So, if a gamma decay occurs for the product of a chemical reaction, we can say that the product is in the excited state and not in the ground state.

So, the product of alpha decay is

(where the star stands for excited)

genius_generous answered 1 year ago

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