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?

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

B.) 1.17

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