QUESTION IMAGE
Question
- which is true of the ^{243}am^{3+} ion?
protons electrons neutrons
(a) 148 148 243
(b) 95 98 243
(c) 95 95 148
(d) 95 92 148
use the mass spectroscopy data below for question 18.
- the figure above represents the mass spectrometry data for an unknown element. which statement best interprets the data below?
(a) there is only one isotope for the unknown element with a mass of 19 amu. this means the isotope has 10 neutrons.
(b) there is only one isotope for the unknown element with a mass of 19 amu. this means the isotope has 9 neutrons.
(c) there is only one ion for the unknown element with a charge of 19 amu. this means the ion has 10 electrons.
(d) there is only one ion for the unknown element with a charge of 19 amu. this means the ion has 9 electrons.
Step1: Determine protons for \(^{243}\text{Am}^{3 +}\)
The atomic number of Americium (\(\text{Am}\)) is \(95\). Protons = atomic number. So protons \(= 95\).
Step2: Calculate electrons for \(^{243}\text{Am}^{3 +}\)
For a cation \(M^{n+}\), electrons \(=\) protons\(-n\). Here \(n = 3\), so electrons \(=95 - 3=92\).
Step3: Calculate neutrons for \(^{243}\text{Am}^{3 +}\)
Mass number \(A=243\). Neutrons \(=A-\) protons. So neutrons \(=243 - 95=148\).
for question 18:
- Mass spectrometry data with a single peak at \(19\) amu implies only one isotope (not ion, as mass is related to isotope mass).
- For an atom, mass number \(A=\) protons + neutrons. If \(A = 19\) and assuming it's a neutral atom (isotope), and if it's \(^{19}\text{F}\) (for example, which is common), protons \(=9\) (atomic number of \(F\)), then neutrons \(=19 - 9=10\).
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D. 95 protons, 92 electrons, 148 neutrons.