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 12 C+ and 14 C+ ions that have the same charge. The 12 C+ ions move in a circle of diameter 25 cm.(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?

Answer 1

A) The diameter of the orbit of 14C+ ions= 29cm

B.) 1.17

Step-by-step explanation:

A)

Given the diameter of circle 12C ions is d12 = 25

Therefore, we now have to apply the second law to an ion moving in a circle. The acceleration is that of uniform circular motion.

a1= v^2/r

= F/m

=\q\vb/ m..... equation (1)

Charge q, speed v, and field B are the same for both ions therefore radius must be directly proportional to the mass.

There r is directly proportional to m

d14/ d12 = r14/ r12

= m14 / m12

To calculate the diameter of the orbit of 14C+ ions, we solve for d14.

d14 is given as

d14= m14/m12 × d12

d14= 14u/ 12u × (25cm)

d14= 29 cm

B.) The frequency is equal to the field divided by the circumference.

F12/ F14 = Circumference 12/ Circumference 14

F12/ F14= (V/ 2 πr12) /(V/ 2 πr14)

=r14/ r12

= m14/ m12

= 14.0u /12.0 u

= 1.17

In conclusion, the ratio of the frequency of revolution of the two types of ion is 1.17