Question

The amount of meat in prehistoric diets can be determined by measuring the ratio of the isotopes ¹⁵N to ¹⁴N in bone from human remains. Carnivores concentrate ¹⁵N, so this ratio tells archaeologists how much meat was consumed. For a mass spectrometer that has a path radius of 12.5 cm for ¹²C ions (mass 1.99 * 10^-26 kg), find the separation of the ¹⁴N (mass 2.32 * 10-26 kg) and ¹⁵N (mass 2.49 * 10-²⁶ kg) isotopes at the detector.

Short answer

The separation of the ¹⁴N and ¹⁵N isotopes at the detector is 2.1357cm

Step-by-step solution

Path radius

Path radius in a magnetic field of a particle is given by

$\mathbf{R}\mathbf{=}\frac{\mathbf{mv}}{\mathbf{qB}}$

Determine the separation

Path radius is given by

$\mathbf{R}\mathbf{=}\frac{\mathbf{mv}}{\mathbf{qB}}$

Therefore, path radius of ¹⁴N is

$\begin{array}{r}{\mathbf{R}}_{{\mathbf{N}}_{\mathbf{14}}}\mathbf{=}\left(\frac{2.32\times {10}^{-26}}{1.99\times {10}^{-26}}\right)\left(\frac{25}{2}\right)\\ \mathbf{=}\mathbf{14}\mathbf{.5729}\mathbf{cm}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

The path of the isotope

$\begin{array}{r}{\mathbf{D}}_{{\mathbf{N}}_{\mathbf{14}}}\\ \mathbf{=}\mathbf{(}\mathbf{14}\mathbf{.5729}\mathbf{)}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\\ \mathbf{=}\mathbf{29}\mathbf{.1457}\mathbf{cm}\end{array}$

Now the radius of the ¹⁵Nis

$\begin{array}{r}{\mathbf{R}}_{{\mathbf{N}}_{\mathbf{15}}}\mathbf{=}\left(\frac{2.49\times {10}^{-26}}{1.99\times {10}^{-26}}\right)\left(\frac{25}{2}\right)\\ \mathbf{=}\mathbf{15}\mathbf{.6407}\mathbf{cm}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the path is

$\begin{array}{r}{\mathbf{D}}_{{\mathbf{N}}_{\mathbf{15}}}\mathbf{=}\mathbf{(}\mathbf{15}\mathbf{.6407}\mathbf{)}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\\ \mathbf{=}\mathbf{31}\mathbf{.2814}\mathbf{cm}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the separation will be

$\begin{array}{r}{\mathbf{D}}_{{\mathbf{N}}_{\mathbf{15}}}\mathbf{-}{\mathbf{D}}_{{\mathbf{N}}_{\mathbf{14}}}\mathbf{=}\mathbf{31}\mathbf{.2814}\mathbf{-}\mathbf{29}\mathbf{.1457}\\ \mathbf{=}\mathbf{2}\mathbf{.1357}\mathbf{cm}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the separation of the ¹⁴N and ¹⁵N isotopes at the detector is 2.1357cm