Chapter 2: Problem 43
Do both members of the following pairs have the same number of protons? Neutrons? Electrons? (a) \({ }_{1}^{2} \mathrm{H}\) and \({ }_{2}^{2} \mathrm{He}\) (b) \({ }_{6}^{14} \mathrm{C}\) and \({ }_{7}^{15} \mathrm{~N}\) (c) \({ }^{10} \mathrm{~F}\) and \({ }_{9}^{18} \mathrm{~F}\) Which pair(s) consist(s) of atoms with the same \(Z\) value? \(N\) value? \(A\) value?
Short Answer
Step by step solution
Identify given values
- Calculate Protons (Z)
- Calculate Neutrons (N)
- Calculate Electrons
Compare Members of Each Pair
Identify Pairs with Same Values
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Protons
For example:
- In the isotope \( \_{1}^{2}H \) (hydrogen), there is 1 proton.
- In \( \_{2}^{2}He \) (helium), there are 2 protons.
- Protons are crucial because they define what element the atom belongs to.
In pairs (a) and (b), the first atoms (\( \_{1}^{2}H \) and \( \_{6}^{14}C \)) have different numbers of protons compared to their pair, showing they are different elements.
Pair (c) shows \( \_{9}^{10}F \) and \( \_{9}^{18}F \) each have 9 protons, indicating these are isotopes of fluorine.
Neutrons
Neutrons stabilize the nucleus, and their count is calculated by subtracting the proton number from the mass number (A):
\( N = A - Z \).
- For \( \_{1}^{2}H \), Neutrons (N) = 2 - 1 = 1.
The number of neutrons alters the mass of the atom without changing its chemical properties.
In our problem:
Pair (b) \( \_{6}^{14}C \) and \( \_{7}^{15}N \) have the same number of neutrons, which is 8.
Electrons
- For Pair (a), \( \_{1}^{2}H \) (1 electron) and \( \_{2}^{2}He \) (2 electrons) do not have the same numbers of electrons.
The difference in their electron count signifies different chemical properties.
In pair (c), \( \_{9}^{10}F \) and \( \_{9}^{18}F \) both have 9 electrons each.Thus, this pair has the same electron count but different neutron numbers, identifying them as isotopes.
Atomic Number (Z)
- For Pair (a), \( \_{1}^{2}H \) and \( \_{2}^{2}He \), Z = 1 and 2 gets different elements.
I pair (c), \( \_{9}^{10}F \) and \( \_{9}^{18}F \), Z = 9. Since they have the same proton count, these are isotopes of fluorine.
Mass Number (A)
\( A = Z + N \).
Distinct isotopes of an element can be identified by their different mass numbers, which reflect the difference in neutron count.
- In Pair (b), \( \_{6}^{14}C \) and \( \_{7}^{15}N \) have distinct mass numbers of 14 and 15, respectively.
This difference highlights that \( \_{6}^{14}C \) has 6 protons and 8 neutrons, while \( \_{7}^{15}N \) has 7 protons and 8 neutrons.
None of our given pairs (a, b, or c) share the same mass number, indicating their distinct identity as different isotopes or elements.