Question

Determining Diet. One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the ${\mathbf{}}^{\mathbf{12}}\mathbf{C}$and ${\mathbf{}}^{\mathbf{13}}\mathbf{C}$isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed , and you want to bend them within a uniform magnetic field in a semicircle of diameter cm for the . The measured masses of these isotopes are $\mathbf{1}\mathbf{.99}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{26}}\mathbf{kg}\left({}^{12}C\right)\mathbf{\text{and}}\mathbf{2}\mathbf{.16}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{26}}\mathbf{kg}\left({}^{13}C\right)$and (a) What strength of magnetic field is required? (b) What is the diameter of the${}^{\mathbf{13}}\mathbf{C}$semicircle? (c) What is the separation of the and ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

Short answer

(a) The strength of magnetic field is required is$\mathrm{B}=8.46\times {10}^{-3}\mathrm{T}$

(b) The diameter of the 13C semicircle is${\mathrm{R}}_{13}=13.6\mathrm{cm}$

(c) The separation of the 12C and 13C ions at the detector at the end of the semicircle is$2\left({\mathrm{R}}_{13}-{\mathrm{R}}_{12}\right)=2.2\mathrm{cm}$ .

Step-by-step solution

Definition of Magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the strength of magnetic field is required and the diameter of the 13C semicircle.

The strength of magnetic field can be calculated as$B=\frac{mv}{\left|q\right|R_{12}}$

Put the all values

$\begin{array}{r}B=\frac{\left(1.99\times {10}^{-26}\mathrm{kg}\right)\left(8.50\times {10}^3\mathrm{m}/\mathrm{s}\right)}{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(0.125\mathrm{m}\right)}\\ B=8.46\times {10}^{-3}\mathrm{T}\end{array}$

Hence, the strength of magnetic field is required is$B=8.46\times {10}^{-3}\mathrm{T}$.

According to the result $\mathrm{R}=\frac{\mathrm{mv}}{\left|\mathrm{q}\right|\mathrm{B}}$Except radius of curvature and mass rest all variables area constant for the ${}^{13}\mathrm{C}{,}^{12}\mathrm{C}$.

So

$\begin{array}{r}\frac{\mathrm{R}}{\mathrm{m}}=\frac{\mathrm{v}}{\left|\mathrm{q}\right|\mathrm{B}}\\ \frac{{\mathrm{R}}_{12}}{{\mathrm{m}}_{12}}=\frac{{\mathrm{R}}_{13}}{{\mathrm{m}}_{13}}\\ {\mathrm{R}}_{13}={\mathrm{R}}_{12}\left(\frac{{\mathrm{m}}_{13}}{{\mathrm{m}}_{12}}\right)\\ \mathrm{Put}\mathrm{all}\mathrm{the}\mathrm{values}\\ {\mathrm{R}}_{13}=(12.5\mathrm{cm})\left(\frac{2.16\times {10}^{-26}\mathrm{kg}}{1.66\times {10}^{-26}\mathrm{kg}}\right)\\ {\mathrm{R}}_{13}=13.6\mathrm{cm}\end{array}$

Hence, the diameter of the 13C semicircle is${\mathrm{R}}_{13}=13.6\mathrm{cm}$.

And the separation of ${}^{12}\mathrm{C},{}^{13}\mathrm{C}$can be calculated as

$\begin{array}{r}2\left({\mathrm{R}}_{13}-{\mathrm{R}}_{12}\right)=2(13.6\mathrm{cm}-12.5\mathrm{cm})\\ 2\left({\mathrm{R}}_{13}-{\mathrm{R}}_{12}\right)=2.2\mathrm{cm}\end{array}$

Hence, the separation of the 12C and 13C ions at the detector at the end of the semicircle is$2\left(R_{13}-R_{12}\right)=2.2\mathrm{cm}$ .