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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?

Respuesta :

Answer:

[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]

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

[tex]\dfrac{q}{m}=\dfrac{E}{BB_0R}[/tex]

[tex]m=\dfrac{qBB_0R}{E}[/tex]

Substituting all the values

[tex]m=\dfrac{1.6\times 10^{-19}\times 0.5\times 0.75\times 02.0232}{1.2\times 10^5}[/tex]

[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]

Mass of Li ions is equal to [tex]m = 1.16 \times 10^{-26}\ Kg[/tex]