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Both barium (Ba) and radium (Ra) are members of Group \(2 \mathrm{~A}\) and are expected to exhibit similar chemical properties. However, \(\mathrm{Ra}\) is not found in barium ores. Instead, it is found in uranium ores. Explain.

Short Answer

Expert verified
Radium is found in Uranium ores and not in Barium ores because it forms as a result of the decay series of Uranium-238. Despite their similar chemical properties, an element's nuclear properties determine where it can be found.

Step by step solution

01

Understand the Decay Series

The uranium decay series is a sequence of radioactive isotopes that are produced as a result of uranium-238 undergoing nuclear decay. The uranium-238 atom undergoes alpha and beta decay and eventually becomes lead-206, which is stable. Along this decay series, several different elements are formed, including radium.
02

Relate Radium with Uranium Decay Series

Radium-226 is actually one of the isotopes that is produced through the decay series of Uranium-238. So, when the uranium ore is decaying, one of the elements it forms along the series is radium.
03

Compare the Behavior of Radium and Barium

Despite being in the same group in the periodic table, barium doesn't form in this decay series. Hence, it's not found in uranium ores. The location where you can find an element is determined by its nuclear, not chemical, properties. This is why we find radium in uranium ores but not barium, even though chemically, they're very similar.

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Most popular questions from this chapter

A radioactive substance undergoes decay as: $$ \begin{array}{cc} \text { Time (days) } & \text { Mass (g) } \\ \hline 0 & 500 \\ 1 & 389 \\ 2 & 303 \\ 3 & 236 \\ 4 & 184 \\ 5 & 143 \\ 6 & 112 \end{array} $$ Calculate the first-order decay constant and the halflife of the reaction.

For each pair of isotopes listed, predict which one is less stable: (a) \({ }_{3}^{6} \mathrm{Li}\) or \({ }_{3}^{9} \mathrm{Li},\) (b) \({ }_{11}^{23} \mathrm{Na}\) or \({ }_{11}^{25} \mathrm{Na},\) (c) \({ }_{20}^{48} \mathrm{Ca}\) or \({ }_{21}^{48} \mathrm{Sc}\).

The radioactive isotope \({ }^{238} \mathrm{Pu},\) used in pacemakers, decays by emitting an alpha particle with a half-life of 86 yr. (a) Write an equation for the decay process. (b) The energy of the emitted alpha particle is \(9.0 \times 10^{-13} \mathrm{~J}\), which is the energy per decay. Assume that all the alpha particle energy is used to run the pacemaker, calculate the power output at \(t=0\) and \(t=10 \mathrm{yr}\). Initially \(1.0 \mathrm{mg}\) of \({ }^{238} \mathrm{Pu}\) was present in the pacemaker (Hint: After \(10 \mathrm{yr}\), the activity of the isotope decreases by 8.0 percent. Power is measured in watts or \(\mathrm{J} / \mathrm{s}\).).

Strontium- 90 is one of the products of the fission of uranium- \(235 .\) This strontium isotope is radioactive, with a half-life of 28.1 yr. Calculate how long (in yr) it will take for \(1.00 \mathrm{~g}\) of the isotope to be reduced to \(0.200 \mathrm{~g}\) by decay.

A radioactive isotope of copper decays as follows: $$ { }^{64} \mathrm{Cu} \longrightarrow{ }^{64} \mathrm{Zn}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=12.8 \mathrm{~h} $$ Starting with \(84.0 \mathrm{~g}\) of \({ }^{64} \mathrm{Cu},\) calculate the quantity of \(^{64}\) Zn produced after 18.4 h.

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