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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 of the electron will be     

  • Option 1)

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

  • Option 2)

    \sqrt{\frac{eV}{m}}

  • Option 3)

    \sqrt{\frac{2eV}{m}}

  • Option 4)

    \frac{2eV}{m}

 

Answers (1)

best_answer

As we hav learnt,

 

when Charged Particle at rest in uniform field -

Kinetiv Energy -

K=\frac{1}{2}mv^{2}=\frac{1}{2}m\left ( \frac{QEt}{m} \right )^{2}=\frac{Q^{2}E^{2}t^{2}}{2m}

K=\frac{1}{2}m\times \frac{2Q\Delta v}{m}=Q\cdot \Delta v

-

 

 Kinetic energy

K = \frac{1}{2}mv^2 = eV \Rightarrow v = \sqrt{\frac{2eV}{m}}

 


Option 1)

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

Option 2)

\sqrt{\frac{eV}{m}}

Option 3)

\sqrt{\frac{2eV}{m}}

Option 4)

\frac{2eV}{m}

Posted by

Aadil

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