A velocity selector consists of electric field E → = E k ^ and magnetic field B → = B j ^ with B = 12 mT . The value E required for an electron of energy 728 eV moving along the positive x -axis to pass undeflected is (Given, mass of electron = 9 . 1 × 10 - 31 kg )

Step-by-step solution

Fiven that E → = E k ^ and B → = 12   j ^   mT Kinetic energy = 728   eV Kinetic energy = 1 2 m v 2 ⇒ 728   eV = 1 2 × 9 . 1 × 10 - 31 × v 2 ⇒ 728 × 1 . 6 × 10 - 19 = 1 2 × 9 . 1 × 10 - 31 × v 2 ⇒ v = 16 × 10 6   m   s - 1 For electron to move undeflected net force on it should be zero. ⇒ e E = e v B ⇒ E = v B = 16 × 10 6 × 12 × 10 - 3 ⇒ E = 192 × 10 3   V   m - 1