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 728eV…

Question

A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ 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 $= 9.1 \times 10^{- 31} kg$ )

MARKS Content Team · Accepted Answer

Correct answer: (A) 192 kV m^(-1). Fiven that E→=Ek^ and B→=12 j^ mT Kinetic energy=728 eV Kinetic energy =1/2mv^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. ⇒eE=evB ⇒E=vB=16×10^6×12×10^(-3) ⇒E=192×10^3 V m^(-1)=192 kV m^(-1)

A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ 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 $= 9.1 \times 10^{- 31} kg$ )

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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 728eV moving along the positive x-axis to pass undeflected is (Given, mass of electron =9.1×10-31 kg)

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