Question

The strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are B=0.500 T and E=1.2x105 V/m. The strength of the magnetic field that seperates the ions is Bo=0.750 T. A stream of single charged Li ions is found to bend in a circular arc of radius 2.32 cm. What is the mass of the Li ions?

Answer 1

`m = 1.16 * 10^(-26)\ Kg`

Step-by-step explanation:

Given,

Magnetic field, B = 0.5 T

Electric field, E = 1.2 x 10⁵ V/m

strength of the magnetic field that separates the ions, Bo=0.750 T

Radius, r = 2.32 cm

Relation of charge to mass ratio is given by

`(q)/(m)=(E)/(BB_0R)`

`m=(qBB_0R)/(E)`

Substituting all the values

`m=(1.6* 10^(-19)* 0.5* 0.75* 02.0232)/(1.2* 10^5)`

`m = 1.16 * 10^(-26)\ Kg`

Mass of Li ions is equal to
`m = 1.16 * 10^(-26)\ Kg`