Question

Two isotopes of carbon, carbon-12 and carbon-13, have masses of 1.993 10-26 kg and 2.159 10-26 kg, respectively. These two isotopes are singly ionized (+e) and each is given a speed of 7.50 105 m/s. The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.6500 T. Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

Answer 1

The spatial separation between the two isotopes after they have traveled through a half cycle is
`1.65* 10^(2)m`

Step-by-step explanation:

The moving of the isotopes must excert centripetal force which is equals to the magnetic force on the ions due to the magnetic.

The centripetal force of the ions can be calculated by the following formula.

`F_(c)=(mv^(2))/(r).............(1)`

Magnetic force on this ions calculated by the following formula.

`F_(m)=qvB.............(2)`

Equate the equations (1) and (2)

`F_(c)=F_(m)`

`(mv^(2))/(r)=qvB`

`v=(mv)/(qB)`

Substitute the values of the both isotopes.

`r_(12)=(1.993* 10^(-26)* 6.13105)/(1.6* 10^(-19)* 0.7700)=9.9041* 10^(-2)m`

`r_(13)=(2.159* 10^(-26)* 6.13105)/(1.6* 10^(-19)* 0.7700)=1.0729* 10^(-1)m`

Now the distance traveled by both isotopes by half circle.

Therefore, the distance between the two isotopes is the diameter of the circle which is equal to the twice the radius.

`Seperation=2*(r_(13)-r_(12))`

`Seperation=2*(1.0729* 10^(-1)-9.9401* 10^(-2))=1.650* 10^(-2)m`

Therefore, The spatial separation between the two isotopes after they have traveled through a half cycle is
`1.65* 10^(2)m`