Question

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, would the velocity of a neutral particle be selected by passage through this device?) The explanation of this is that the mass and the charge control the resolution of the device--particles with the wrong velocity will be accelerated away from the straight line and will not pass through the exit slit. If the acceleration depends strongly on the velocity, then particles with just slightly wrong velocities will feel a substantial transverse acceleration and will not exit the selector. Because the acceleration depends on the mass and charge, these influence the sharpness (resolution) of the transmitted particles.

Assume that you want a velocity selector that will allow particles of velocity v to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with __________.

a. both q and m large
b. q large and m small
c. q small and m large
d. both q and m small

Answer 1

Answer:

The answer is (b). q large and m small.

Explanation:

We know that whenever a charged particle having charge '$q$' is moving with a velocity '$v$' through an electric field ($\overrightarrow{E}$) and a magnetic field ($\overrightarrow{B}$), it will experience a force, called Lorentz Force ($F_{L}$), given by

$\overrightarrow{F_{L}} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B} )$

Now, due to this Lorentz force if the charge having mass '$m$' gains an acceleration '$a$', then

$&&ma = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})\\&or,& a = \frac{q}{m}(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$

The deviation of the particle will be maximum when the acceleration of the article is maximum. Therefore to obtain the best possible resolution, according to above equation, '$q$' has to be large and '$m$' has to be small.