In an electron gun, electrons are accelerated through a potential difference of $V$ volt. Taking electronic charge and mass to be $e$ and $m$ respectively, the maximum velocity attained by them is the :

\frac{2 e V}{m}

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

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\frac{2 e}{m}

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

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Solution

The correct option is B

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

Kinetic energy of electron accelerated through a potential

V= eV

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In an electron gun, electrons are accelerated through a potential difference of V volt. Taking electronic charge and mass to be e and m respectively, the maximum velocity attained by them is the :

The correct option is B √2eVmKinetic energy of electron accelerated through a potentialV= eV

In an electron gun, electrons are accelerated through a potential difference of $V$ volt. Taking electronic charge and mass to be $e$ and $m$ respectively, the maximum velocity attained by them is the :

The correct option is B $\sqrt{\frac{2 e V}{m}}$
Kinetic energy of electron accelerated through a potential
V= eV