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

Question

An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed will be (A)  (2eV/m)  (B)  V√(e/m)  (C)  √(3eV/m)  (D)  √(2eV/m)

Tardigrade Admin · Accepted Answer

Kinetic energy of electron = eV ⇒ (1/2)mv2=eV ⇒ v=√(2eV/m)

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

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

$V \frac{e}{m}$

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

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

Kinetic energy of electron = eV $\Rightarrow$ $\frac{1}{2} m v^{2} = e V$ $\Rightarrow$ $v = \frac{2 e V}{m}$

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

m2eV​

Vme​​

m3eV​​

m2eV​​

Kinetic energy of electron = eV ⇒ 21​mv2=eV ⇒ v=m2eV​​

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

$V \frac{e}{m}$

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

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

Kinetic energy of electron = eV $\Rightarrow$ $\frac{1}{2} m v^{2} = e V$ $\Rightarrow$ $v = \frac{2 e V}{m}$