1 Answer

Rory Shaw · Answer 1 · 2023-09-10T02:03:18+0000

The question asks us to identify the element formed when americium-241 (Am-241) emits an alpha particle.

Considering the alpha radiation, a particle containing 4 protons and 2 neutrons is emitted:

$\text{Alpha particle }\rightarrow^4_2He$

We can write the alpha decay of Am-241 as:

$^(241)_(95)Am\text{ }\rightarrow^4_2He+^{_{}237}_(93)Np$

which means that americium-241 (Am-241) lost an alpha particle (He-4) and formed neptunium-237 (Np-237).

Therefore, the atomic symbol of the nuclide daughter formed when americium-241 undergoes an alpha decay is Np (option A).

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