Explanation Given Info: The path radius ( R 12 ) for 12 C ions is 12.5 cm . The mass ( m 12 ) of the ion 12 C is 1.99 × 10 − 26 kg . The mass ( m 14 ) of the ion 14 N is 2.32 × 10 − 26 kg . The mass ( m 15 ) of the ion 15 N is 2.49 × 10 − 26 kg . Formula to calculate the radius of the path of the ion is, R = m v | q | B m is the mass of the ion, v is the velocity of the ion, | q | is the magnitude of the charge of the ion, and B is the magnetic field in which the ion is passing through. Rearranging the equation, R m = v | q | B The ions 12 C , 14 N , and 15 N has same value of velocity ( v ) , same amount of charge ( | q | ) , and same value of the magnetic field ( B ) when it passes through the mass spectrometer. Formula to calculate the radius of the path of the ion 12 C is, R 12 = m 12 v | q | B (I) m 12 is the mass of the ion 12 C . Formula to calculate the radius of the path of the ion 14 N is, R 14 = m 14 v | q | B (II) m 14 is the mass of the ion 14 N . Formula to calculate the radius of the path of the ion 15 N is, R 15 = m 15 v | q | B (III) m 15 is the mass of the ion 15 N . Dividing the equation (II) by the equation (I), R 14 R 12 = m 14 v | q | B m 12 v | q | B R 14 R 12 = m 14 m 12 R 14 = R 12 ( m 14 m 12 ) Substitute 12

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Chapter 27, Problem 27.33E

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The separation of 14N and 15N isotopes at the detector.

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Chapter 27, Problem 27.32E

Chapter 27, Problem 27.34E

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Chapter 27 Solutions

University Physics with Modern Physics (14th Edition)

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Chapter 27 Solutions