Question

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 154 V/m and the magnetic field is 3.18×10-2 T. The ions next enter a uniform magnetic field of magnitude 1.75×10-2 T that is oriented perpendicular to their velocity.
a)How fast are the ions moving when they emerge from the velocity selector?
v=____________m/s

b)If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?
m=__________kg

Answer 1

Answer:

a. v  = 4748.42 m/s

b. mass = 10.32 * 10⁻²⁶kg

Explanation:

Electric field, E = 151 V/m

Magnetic field , B =3.18×10⁻²T

(a)

When the ion emerges from the velocity selector, its velocity is  :

             v = E / B................(1)

                = (151 V/m) / (3.18×10 ⁻² T)

          velocity, v  = 4748.42 m/s

(b)

Magnetic field B =1.75×10⁻² T

radius of the path of the ions r = 17.5 cm = 0.175 m

magnetic force, F = Bvq   ................... (1)

Applying Newton's second law of motion ,

     Force F = ma  

here , centripetal acceleration a = v2/r

     force F = mv2/r   ................ (2)  

comparing equations (1) & (2) , we have,

        Bvq = mv2/r  

        Bq = mv/r  

mass, m = Bqr/v

            = (1.75×10⁻² T)(1.6*10⁻¹⁹ C)(0.175 m) /(4748.42 m/s)

       mass = 10.32 * 10⁻²⁶kg