The electrons in a particle beam each have a kinetic energy of 1-6 - 10-7 J- What are the magnitude and direction of the electric field that stops these electrons in a distance of 10-0 cmA- 106 v - m perpendicular to velocity of electronsB- 103 v - m perpendicular to velocity of electronsC- 103 v - m in the direction of velocity of electronsD- 103 v - m positive direction of velocity of electrons

The correct option is C 10^3 v/m in the direction of velocity of electrons W = FS 1/2mv^2=EqS 1.6 × 10^ 17=E × 1.6 × 10^ 19× 10 × 10^ 2 E = 10^3 V/m in the ...

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Byju's Answer

Standard XII

Chemistry

Thomson's Discovery of Electrons

The electrons...

Question

The electrons in a particle beam each have a kinetic energy of $1.6 \times 10^{- 7} J$ . What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0cm

10^{3}

v/m in the direction of velocity of electrons

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10^{3}

v/m positive direction of velocity of electrons

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10^{3}

v/m perpendicular to velocity of electrons

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10^{6}

v/m perpendicular to velocity of electrons

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Solution

The correct option is A $10^{3}$ v/m in the direction of velocity of electrons
W = FS
$\frac{1}{2} m v^{2} = E q S$
$1.6 \times 10^{- 17} = E \times 1.6 \times 10^{- 19} \times 10 \times 10^{- 2}$
$E = 10^{3} V / m$ in the direction of velocity of electrons

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The electrons in a particle beam each have a kinetic energy of 1.6×10−7J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0cm

The correct option is A 103 v/m in the direction of velocity of electrons W = FS 12mv2=EqS 1.6×10−17=E×1.6×10−19×10×10−2 E=103 V/m in the direction of velocity of electrons

The electrons in a particle beam each have a kinetic energy of $1.6 \times 10^{- 7} J$ . What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0cm

The correct option is A $10^{3}$ v/m in the direction of velocity of electrons
W = FS
$\frac{1}{2} m v^{2} = E q S$
$1.6 \times 10^{- 17} = E \times 1.6 \times 10^{- 19} \times 10 \times 10^{- 2}$
$E = 10^{3} V / m$ in the direction of velocity of electrons