Question

Write the nuclear equation to describe the alpha decay of 243 95 Am

Answer 1

Here's what I get.

Step-by-step explanation:

Your unbalanced nuclear equation is:

`_(95)^(243)\text{Am} \longrightarrow \, _(x)^(y)\text{Z} +\, _(2)^(4)\text{He}`

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

95 = x + 2, so x = 95 - 2 = 93

243 = y + 4, so y = 243 - 4 = 239

Element 93 is neptunium, so the nuclear equation becomes

`_(95)^(243)\text{Am} \longrightarrow \, _(93)^(239)\text{Np} + \, _(2)^(4)\text{He}`

Answer 2

Answer: The nuclear reaction is written below.

Step-by-step explanation:

Alpha decay is defined as the decay process in which alpha particle is released. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units and a mass of 4 units.

The released alpha particle is also known as helium nucleus.

In this decay process, the atomic number of the atom decreases by 2 units and the mass number decreases by 4 units.

The chemical equation for alpha decay process follows:

`_Z^A\textrm{X}\rightarrow _(Z-2)^(A-4)\textrm{Y}+_2^4\alpha`

The chemical equation for alpha decay of
`_(95)^(243)\textrm{Am}` follows:

`_(95)^(243)\textrm{Am}\rightarrow _(93)^(239)\textrm{Np}+_(2)^(4)\alpha`

Hence, the nuclear reaction is written above.