## In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to dis

Question

In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish 12C and 14C ions. The 12C ions move in a circle of diameter 42.3 cm. Use these atomic mass values: 12C, 12.0 u; 14C, 14.0 u.

(a) What is the diameter of the orbit of 14 C+ ions?

(b) What is the ratio of the frequencies of revolution for the two types of ion?

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1 year 2021-08-14T06:03:32+00:00 1 Answers 7 views 0

Explanation:

(a) For this problem we can use the expression

$$v=\frac{qBr}{m}$$

where B is the magnetic field, r is the radius of the curve described by the ion, m is the mass and q is the electric charge.

In this case, both isotopes have the same velocity, and we can assume that both isotopes has been ionized to have the same charge. Being r and r’ the radius of the curve for 12C ion and 14C ion respectively, we have

$$v=v’\\\frac{qBr}{m}=\frac{qBr’}{m’}\\r’=\frac{m’r}{m}=\frac{(14u)(21.15cm)}{(12u)}=24.67cm\\$$

Hence, 2*24.67cm=49.35cm is the diameter of the orbit for 14C.

(b) we can calculate ω by using

$$\omega_{12C} =\frac{v}{r}\\\omega_{14C}=\frac{v}{r’}\\$$

By dividing these expressions we have

$$\frac{\omega_{12C}}{\omega_{14C}}=\frac{r’}{r}=\frac{24.67cm}{21.15cm}=1.16$$

The isotope 12C has 1.16 more angular frecuency than 14C

I hope this is useful for you

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