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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 12 C and 13 C isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 km/s, and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the 12 C. The measured masses of these isotopes are 1.99×10−26kg(12C) and 2.16×10−26kg(13C).
(a) What strength of magnetic field is required?
(b) What is the diameter of the 13 C semicircle?
(c) What is the separation of the 12 C and 13 C ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

Answer :

boffeemadrid

Answer:

[tex]0.0084575\ \text{T}[/tex]

[tex]0.272\ \text{m}[/tex]

2.2 cm easily observable

Explanation:

[tex]m_1[/tex] = Mass of 12 C = [tex]1.99\times 10^{-26}\ \text{kg}[/tex]

[tex]m_2[/tex] = Mass of 13 C = [tex]2.16\times 10^{-26}\ \text{kg}[/tex]

[tex]r_1[/tex] = Radius of 12 C = [tex]\dfrac{25}{2}=12.5\ \text{cm}[/tex]

B = Magnetic field

v = Velocity of atom = 8.5 km/s

[tex]r_2[/tex] = Radius of 13 C

The force balance of the system is

[tex]qvB=\dfrac{m_1v^2}{r}\\\Rightarrow B=\dfrac{m_1v}{rq}\\\Rightarrow B=\dfrac{1.99\times 10^{-26}\times 8500}{12.5\times 10^{-2}\times 1.6\times 10^{-19}}\\\Rightarrow B=0.0084575\ \text{T}[/tex]

The required magnetic field is [tex]0.0084575\ \text{T}[/tex]

Radius is given by

[tex]r=\dfrac{mv}{qB}[/tex]

[tex]r\propto m[/tex]

So

[tex]\dfrac{r_2}{r_1}=\dfrac{m_2}{m_1}\\\Rightarrow r_2=\dfrac{m_2}{m_1}r_1\\\Rightarrow r_2=\dfrac{2.16\times 10^{-26}}{1.99\times 10^{-26}}\times 12.5\times 10^{-2}\\\Rightarrow r_2=0.136\ \text{m}[/tex]

The required diameter is [tex]2\times 0.136=0.272\ \text{m}[/tex]

Separation is given by

[tex]2(r_2-r_1)=2(0.136-0.125)=0.022\ \text{m}[/tex]

The distance of separation is 2.2 cm which is easily observable.

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