A beam of alpha particles passes undeflected through a velocity selector containing crossed electric and magnetic fields. The electric field is 4 kV m ⁻¹ and the magnetic field is 20 mT . The beam then enters a second region containing a uniform magnetic field of 50 mT perpendicular to the velocity of the particles, where it moves in a circular path of radius 0.8 m . If the charge of an alpha particle is 3.2 10⁻¹⁹ C

Step-by-step solution

For the alpha particles to pass undeflected through the velocity selector, the electric and magnetic forces must balance: v = E B₁ Given E = 4 kV m ⁻¹ = 4 10³ V m ⁻¹ and B₁ = 20 mT = 20 10⁻³ T : v = 4 10³ 20 10⁻³ = 2 10⁵ m s ⁻¹ In the second region, the magnetic field B₂ provides the necessary centripetal force for circular motion: R = mv qB₂ Rearranging for mass m : m = qB₂R v Substitute the given values ( q = 3.2 10⁻¹⁹ C , B₂ = 50 10⁻³ T , R = 0.8 m ): m = 3.2 10⁻¹⁹ 50 10⁻³ 0.8 2 10⁵ m = 12.8 10⁻²² 2 10⁵ = 6.4 10