An electron of mass M kg and charge coulomb travels from rest through a potential difference of V volt. The final, y velocity of the electron is (in m7s)

Question

An electron of mass M kg and charge coulomb travels from rest through a potential difference of V volt. The final, y velocity of the electron is (in m7s) (A) (2eV/M) (B) (2MV/e) (C) √(2eV/M) (D) √(2MV/e)

Tardigrade Admin · Accepted Answer

Energy gained by an electron = q Δ V = V × e this energy is in the from kinetic energy ⇒ (1/2) Mv f 2= V × e ⇒ v f 2=(2 eV / M ) ⇒ v f =√( 2 e V / M )

Q. An electron of mass M kg and charge coulomb travels from rest through a potential difference of V volt. The final, y velocity of the electron is (in m7s)

$\frac{2 e V}{M}$

$\frac{2 M V}{e}$

$\frac{2 e V}{M}$

$\frac{2 M V}{e}$

Energy gained by an electron
$= q Δ V$ $= V \times e$
this energy is in the from kinetic energy
$\Rightarrow \frac{1}{2} M v_{f}^{2} = V \times e$
$\Rightarrow v_{f}^{2} = \frac{2 e V}{M}$
$\Rightarrow v_{f} = \frac{2 e V}{M}$

Q. An electron of mass M kg and charge coulomb travels from rest through a potential difference of V volt. The final, y velocity of the electron is (in m7s)

M2eV​

e2MV​

M2eV​​

e2MV​​

Energy gained by an electron =qΔV =V×e this energy is in the from kinetic energy ⇒21​Mvf2​=V×e ⇒vf2​=M2eV​ ⇒vf​=M2eV​​

$\frac{2 e V}{M}$

$\frac{2 M V}{e}$

$\frac{2 e V}{M}$

$\frac{2 M V}{e}$

Energy gained by an electron
$= q Δ V$ $= V \times e$
this energy is in the from kinetic energy
$\Rightarrow \frac{1}{2} M v_{f}^{2} = V \times e$
$\Rightarrow v_{f}^{2} = \frac{2 e V}{M}$
$\Rightarrow v_{f} = \frac{2 e V}{M}$