The velocity acquired by an electron at rest | vedclass

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(D) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K.E. = eV$. Since the electron starts from re

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$8000$

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(D) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K.E. = eV$. Since the electron starts from rest,the kinetic energy is equal to $\frac{1}{2}mv^2$. Equating the two: $\frac{1}{2}mv^2 = eV$. Solving for velocity $v$: $v = \sqrt{\frac{2eV}{m}}$. Given: $e = 1.6 	imes 10^{-19} \ C$,$V = 180 \ V$,$m = 9 	imes 10^{-31} \ kg$. Substituting the values: $v = \sqrt{\frac{2 	imes 1.6 	imes 10^{-19} 	imes 180}{9 	imes 10^{-31}}}$. $v = \sqrt{\frac{576 	imes 10^{-19}}{9 	imes 10^{-31}}} = \sqrt{64 	imes 10^{12}} = 8 	imes 10^6 \ m/s$. Converting to $km/s$: $v = 8000 \ km/s$.

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)

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