An electron is accelerated through a potential difference of 1000 volt. The velocity acquired by the electron is

2.0 \times 10^{8} m / s^{- 1}

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1.88 \times 10^{7} m / s^{- 1}

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1.5 \times 10^{7} m / s^{- 1}

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1.5 \times 10^{8} m / s^{- 1}

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Solution

The correct option is B $1.88 \times 10^{7} m / s^{- 1}$
$q V = \frac{1}{2} m v^{2}$
$v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 10^{3}}{9 \times 10^{- 31}}} = \sqrt{3.55 \times 10^{- 16 + 30}}$
$v = 1.88 \times 10^{7} m / s$

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An electron is accelerated through a potential difference of 1000 volt. The velocity acquired by the electron is

The correct option is B 1.88×107m/s−1qV=12mv2 v=√2eVm=√2×1.6×10−19×1039×10−31=√3.55×10−16+30 v=1.88×107m/s

The correct option is B $1.88 \times 10^{7} m / s^{- 1}$
$q V = \frac{1}{2} m v^{2}$
$v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 10^{3}}{9 \times 10^{- 31}}} = \sqrt{3.55 \times 10^{- 16 + 30}}$
$v = 1.88 \times 10^{7} m / s$