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The amount of meat in prehistoric diets can be determined by measuring the ratio of the...

The amount of meat in prehistoric diets can be determined by measuring the ratio of the isotopes nitrogen-15 to nitrogen-14 in bone from human remains. Carnivores concentrate 15 N + 15N+ , so this ratio tells archaeologists how much meat was consumed by ancient people. Suppose you use a velocity selector to obtain singly-ionized (missing one electron) isotopes of speed 8.50 km/s km/s . The spectrometer is calibrated by where 12 C + 12C+ arrive after traveling around a semicircle. Suppose the 12 C + 12C+ ions travel along a semicircle with diameter 30.0 cm cm . Use the actual masses of these ions in your calculations: 1.99×10−26 kg kg for 12 C + 12C+ 2.32×10−26 kg kg for 14 N + 14N+ 2.49×10−26 kg kg for 15 N +

Find the separation of the 14 N + 14N+ and 15 N + 15N+ isotopes at the detector in centimeters.

Solutions

Expert Solution

Step 1: Find value of uniform magnetic field, using force balance on the 12C atoms:

Force on Carbon-12 ions will be:

Force due to circular motion = Force due to magnetic field

Fc = Fm

m1*V^2/R1 = q*V*B

B = m1*V/(q*R1)

m1 = mass of Carbon-12 = 1.99*10^-26 kg

V = Speed of ions = 8.50 km/sec = 8500 m/s

q = charge on simgly-ionized atoms = +e = 1.6*10^-19 C

R1 = radius of circular path of Carbon-12 = diameter/2 = 30/2 = 15 cm = 0.15 m

So,

B = 1.99*10^-26*8500/(1.6*10^-19*0.15)

B = 7.048*10^-3 T = Uniform magnetic field

Step 2: Find out the radius of path followed by Nitrogen-14:

Since we know that:

m2*V^2/R2 = q*V*B

R2 = m2*V/(q*B)

m2 = mass of Nitrogen-15 = 2.32*10^-26 kg

R2 = 2.32*10^-26*8500/(1.6*10^-19*7.048*10^-3)

R2 = 0.1749 m = 17.49 cm

Step 3: Find out the radius of path followed by Nitrogen-15:

Since we know that:

m2*V^2/R2 = q*V*B

R2 = m2*V/(q*B)

m2 = mass of Nitrogen-15 = 2.49*10^-26 kg

R2 = 2.49*10^-26*8500/(1.6*10^-19*7.048*10^-3)

R2 = 0.1877 m = 18.77 cm

Step 4: diameter of semicircle path of Nitrogen-14 will be:

d1 = 17.49*2 = 34.98 cm

diameter of semicircle path of Nitrogen-15 is:

d2 = 18.77*2 = 37.54 cm

So separation between 14N and 15N isotopes will be:

d = d2 - d1 = 37.54 - 34.98

d = 2.56 cm

Let me know if you've any query.

genius_generous answered 1 year ago
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