We’re being asked to **propose a reason for the unusual stability of isotope Q given the nuclear reaction**:

${}_{\mathbf{97}}{}^{\mathbf{249}}\mathbf{Bk}\mathbf{+}\mathbf{Q}\mathbf{\to}\mathbf{Z}\mathbf{\to}{}_{\mathbf{117}}{}^{\mathbf{294}}\left[117\right]\mathbf{+}\mathbf{3}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathbf{n}$

Recall that the ** stability of an isotope** depends on the ratio of neutrons to protons (N/Z). This means for stable isotopes with:

• **Z ≤ 20**; the N/Z ratio should be **equal to 1.0**

• **20 < Z ≤ 40**; the N/Z ratio should be **equal to 1.25**

• **40 < Z ≤ 80**; the N/Z ratio should be **equal to 1.50**

• **Z < 83**; the stable nuclide **does not exist**

The N/Z values for stable isotopes form the ** valley or band of stability**:

We shall determine the identity of isotope Q:

We are to identify what Z is first:

Consider the part of the equation as follows:

$\mathbf{Z}\mathbf{\to}{}_{\mathbf{117}}{}^{\mathbf{294}}\mathbf{\left[}\mathbf{117}\mathbf{\right]}\mathbf{+}\mathbf{3}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathbf{n}$

Recall:

**Step 1:** Let’s first balance the atomic mass:

x = 294 + 3

x = **297** = atomic mass of the unknown isotope

**Step 2:** Next, we balance the atomic number:

x = 117+0

x = **117** = atomic number of the unknown istope

**Step 3:**** **The identity of Z is:

${}_{\mathbf{117}}{}^{\mathbf{297}}\mathbf{Z}$

We are to identify what Q is:

Consider the part of the equation as follows:

${}_{\mathbf{97}}{}^{\mathbf{249}}\mathbf{Bk}\mathbf{+}\mathbf{Q}\mathbf{\to}{}_{\mathbf{117}}{}^{\mathbf{297}}\mathbf{Z}$

**Step 1:** Let’s first balance the atomic mass:

249 + x = 297

x =297-249 = **48** = atomic mass of the unknown isotope

**Step 2:** Next, we balance the atomic number:

97 + x = 117

x =117-97 = **20** = atomic number of the unknown istope

**Step 3:**** **The identity of Q (i.e. the atomic number determines the identity of the element) is:

${}_{\mathbf{20}}{}^{\mathbf{48}}\mathbf{Ca}$

In 2010, a team of scientists from Russia and the U.S. reported the creation of the first atom of element 117, which is not yet named and is denoted [117]. The synthesis involved the collision of a target of _{97}^{249}Bk with accelerated ions of an isotope which we will denote Q. The product atom, which we will call Z, immediately releases neutrons and forms _{117}^{294}[117]:_{97}^{249}Bk + Q → Z → _{117}^{294}[117]+
3_{0}^{1}n

Isotope Q is unusual in that it is very long-lived (its half-life is on the order of 10^{19} yr) in spite of having an unfavorable neutron-to-proton ratio.

Can you propose a reason for its unusual stability?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Band of Stability concept. You can view video lessons to learn Band of Stability. Or if you need more Band of Stability practice, you can also practice Band of Stability practice problems.