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

Answer:

$m = 1.16 \times 10^{-26}\ Kg$

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

$\dfrac{q}{m}=\dfrac{E}{BB_0R}$

$m=\dfrac{qBB_0R}{E}$

Substituting all the values

$m=\dfrac{1.6\times 10^{-19}\times 0.5\times 0.75\times 02.0232}{1.2\times 10^5}$

$m = 1.16 \times 10^{-26}\ Kg$

Mass of Li ions is equal to $m = 1.16 \times 10^{-26}\ Kg$