In a mass spectrometer ions are accelerated through a potential difference V and

Question

In a mass spectrometer ions are accelerated through a potential difference V and enter a region of uniform magnetic field strength B as shown in the figure. The magnetic force causes the ions to move through a semicircular path where they are observed by a detector. If y is the lateral displacement of the ion, determine the charge-to-mass ratio in terms of y, B and V.

In a mass spectrometer ions are accelerated through a potential difference V and enter a region of uniform magnetic field strength B as shown in the figure. The magnetic force causes the ions to move through a semicircular path where they are observed by a detector. If y is the lateral displacement of the ion, determine the charge-to-mass ratio in terms of y, B and V.

Explanation / Answer

Potential diff. = V volts

Magnetic field strength = B

displacement of ion = y

KE of the particle = 1/2 m v2and its produced by the voltage qV

1/2 m v2= V which gives us

v = sqrt (2q V/m) eq(1)

We know that,

F(Lorentz) = q v B and F(centripital)= m v2 / y (v is velocity of the particle)

which gives us, y = mv/qB eq(2)

putting the value of v frm eq(1) to eq(2)

y = 1/B sqrt (2mV/q)

squaring both sides

y2 = 1/B2 (2mV/q)

This gives us,

q/m = 2V / (B y)2 (charge to mass ratio)

or m/q = (By)2 / (2V) (mass to charge ratio)

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