Some Properties of Atoms — Question 3

Section 40-1 · Physics pages 6576–6577

Question 3 An isotopic abundance can be inferred from a mean atomic mass A hypothetical element has only isotopes of masses 62.0 u and 65.0 u. Its measured mean atomic mass is 63.2 u. Treat the quoted masses as exact in this model. 62 u 65 u 1 −x x Schematic; labels, not drawing lengths, specify the data. (a) Find the number fraction of the heavier isotope. (b) Find its mass fraction in a macroscopic sample.
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Question 3 - Solution An isotopic abundance can be inferred from a mean atomic mass 62 u 65 u 1 −x x (a) Method and calculation ¯m = (1 −x)ml + xmh = ml + x(mh −ml). x = (63.2 −62.0)/(65.0 −62.0) = 0.400 . (b) Interpretation and further calculation wh = Nxmh N ¯m = xmh ¯m . wh = 26.0/63.2 = 0.4114 . Physical check Number fractions weight the mean mass; mass fractions weight the sample mass. The heavier isotope has a larger mass fraction than number fraction.

Question 3

An isotopic abundance can be inferred from a mean atomic mass

A hypothetical element has only isotopes of masses 62.0u62.0\,\mathrm u and 65.0u65.0\,\mathrm u. Its measured mean atomic mass is 63.2u63.2\,\mathrm u. Treat the quoted masses as exact in this model.

Diagram shown in the original SVG worksheet.

  1. Find the number fraction of the heavier isotope.

  2. Find its mass fraction in a macroscopic sample.

Question 3 - Solution

An isotopic abundance can be inferred from a mean atomic mass

Diagram shown in the original SVG worksheet.

(a) Method and calculation

m‾=(1−x)ml+xmh=ml+x(mh−ml).\bar m=(1-x)m_l+xm_h=m_l+x(m_h-m_l).x=(63.2−62.0)/(65.0−62.0)=0.400.\boxed{x=(63.2-62.0)/(65.0-62.0)=0.400}.

(b) Interpretation and further calculation

wh=NxmhNm‾=xmhm‾.w_h=\frac{Nxm_h}{N\bar m}=\frac{xm_h}{\bar m}.wh=26.0/63.2=0.4114.\boxed{w_h=26.0/63.2=0.4114}.

Physical check

Number fractions weight the mean mass; mass fractions weight the sample mass. The heavier isotope has a larger mass fraction than number fraction.