An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed will be :

\frac{2 e V}{m}

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V \sqrt{\frac{e}{m}}

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\sqrt{\frac{3 e V}{m}}

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\sqrt{\frac{2 e V}{m}}

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Solution

The correct option is D $\sqrt{\frac{2 e V}{m}}$
Energy imparted by electric field will be reflected in the increase
in kinetic energy.
so , $\frac{1}{2} m v^{2} = e V$
$v = \sqrt{\frac{2 e V}{m}}$

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An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed will be :

The correct option is D √2eVmEnergy imparted by electric field will be reflected in the increasein kinetic energy. so , 12mv2=eVv=√2eVm

The correct option is D $\sqrt{\frac{2 e V}{m}}$
Energy imparted by electric field will be reflected in the increase
in kinetic energy.
so , $\frac{1}{2} m v^{2} = e V$
$v = \sqrt{\frac{2 e V}{m}}$