Mass Spectrometer Charge Ratio

Mass Spectrometer Charge Ratio (2007)

Question: In a mass spectrometer with constant \( V \) and \( B \), the ratio proportional to the charge \( q \) (for fixed ion mass) is:

(a) \( \frac{1}{R^2} \)

(b) \( R^2 \)

(c) \( R \)

(d) \( \frac{1}{R} \)

Solution

Step 1: Kinetic Energy from Electric Potential
\[ qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}} \]

Step 2: Magnetic Force as Centripetal Force
\[ qvB = \frac{mv^2}{R} \implies R = \frac{mv}{qB} \]

Step 3: Substitute \( v \) into Radius Formula
\[ R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2mV}{qB^2}} \] Square both sides: \[ R^2 = \frac{2mV}{qB^2} \implies q \propto \frac{1}{R^2} \quad (\text{for fixed } m, V, B) \]

Answer: Option (a) \( \frac{1}{R^2} \) is correct.