Question

A velocity selector in a mass spectrometer uses a 0.134 T magnetic field. What electric field strength is needed to select a speed of 4.67 × 10^6 m/s?

a) 1.25 V/m

b) 2.50 V/m

c) 3.75 V/m

d) 5.00 V/m

Answer 1

To select a particle speed of 4.67 × 10¶ m/s in a 0.134 T magnetic field, the needed electric field strength in the mass spectrometer's velocity selector is 626 V/m.

Step-by-step explanation:

In a mass spectrometer, a velocity selector permits only particles with a certain velocity to pass through undeflected when both the magnetic field and electric field are applied. The required strength of the electric field (E) to select a certain speed (v) is calculated using the balance between the magnetic force (qvB) and the electric force (qE) exerted on a charged particle. Therefore, to find the necessary electric field strength, we set the magnetic force equal to the electric force (qE = qvB), and since the charge (q) cancels out, E can be found simply by E = vB.

For a velocity selector in a mass spectrometer using a 0.134 T magnetic field to select a speed of 4.67 × 106 m/s, the electric field strength needed is calculated as:

E = vB = (4.67 × 106 m/s)(0.134 T) = 625.98 V/m, which can typically be rounded to 626 V/m depending on significant figures considerations.