2.In a mass spectrograph, a singly charged ion is accelerated across a 2030 volt potential difference. It then enters a magnetic field of 0.500 Teslas. The radius of the obit was measured to be 12.0 cm. What is the mass of the ion

m = 1.42 × 10^(-25) kg

Explanation:

We are given;

Potential Difference; V = 2030 V

Magnetic field; B = 0.5 T

Radius; r = 12 cm = 0.12 m

Now, if the mass of the particle is m and the electron charge is q, and considering the fact that the magnetic field will provide a centripetal motion, then from magnetic field formula, force is;

F = mv²/r = qvB

Thus;

mv²r = qvB

v = qBr/m

Where v is velocity.

Now, the electrical energy will be equal to the kinetic energy.

Thus;

qV = ½mv²

Where V is potential difference

Thus;

v = √(2qV/m)

Thus, equating both v equations to get;

qBr/m = √(2qV/m)

Square both sides to get;

q²B²r²/m² = 2qV/m

Simplifying gives;

m = qB²r²/2V

q is charge of one electron and has a value of 1.6 × 10^(-19) C

Thus;

m = (1.6 × 10^(-19) × 0.5² × 0.12²)/(2 × 2030)

m = 1.42 × 10^(-25) kg

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