onstants Part A The amount of meat in prehistoric diets can be determined by mea

Question

onstants Part A 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 concentrateN, 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) nitrogen atoms of speed 8.65 km/s semicircle within a uniform magnetic field. The 14N atoms travel along a semicircle with a diameter of 22.5 cm . The measured masses of these isotopes are 2.32x10-26 kg (14N) and 2.49x1026 kg for (15N) Find the separation of the 14N and 15N isotopes at the detector Express your answer in centimeters and bend them along a cm Submit Request Answer

Explanation / Answer

Let the mass of the isotope 15N is M = 2.49 × 10-26 kg

Let the mass of the isotope 14N is m = 2.32 × 10-26 Kg

Since singly ionized , charge on them is q = 1.6 × 10-19 C

Velocity of this singly ionized nitrogen atoms is v = 8.65 km/s

= 8.65 × 103 m/s

Let the uniform magnetic field applied be B

Due to this field , ionized atoms start to move in circular path.

But we trace them after completing a semi circle to calculate the maximum seperation between the two ions.

Radius of this semi circle is same as the circular path.

To calculate the radius of the semicircle traced by 15N atoms , we need the magnitude of the magnetic field B. We calculate B using diameter of semicircle traced by 14N atoms.

Diameter of the semi circle traced by 14N atoms is d1 = 22.5 cm

Radius is r1 = d1 /2 = 22.5 / 2 = 11.25 cm = 0.1125 m

Radius of path is given by r1 = mv / qB

B = mv / q r1

= ( 2.32 × 10-26 × 8.65 × 103 ) / ( 1.6 × 10-19 × 0.1125 )

= ( 20.06 × 10-4 ) / 0.18

= 111.48 × 10-4 T

= 0.011 T

Radius of the semicircle traced by ions of 15N is given by

r2 = M v / q B

= ( 2.49 × 10-26 × 8.65 × 103 ) / ( 1.6 × 10-19 × 0.011 )

= ( 21.53 × 10-23 ) / ( 0.0176 × 10-19 )

= 1223.77 × 10-4

= 12.23 × 10-2 m

= 12.23 cm

Diameter of the semicircle is d2 = 2 r2 = 24.46 cm

We can observe that radius of the semicircle traced by ion with higher mass is greater.

This is evident as v , q and B are constant for both the isotopes.

Seperation between the two isotopes is equal difference of their diameters.

S = d2 - d1 = 24.46 - 22.50 = 1.96 cm

So seperation between the two isotopes is S = 1.96 cm

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