An electron is accelerated through potential difference of 45.5 volt. The velocity acquired by it is (in ${m s}^{- 1}$ ) :

4 \times 10^{6}

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4 \times 10^{- 6}

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10^{6}

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Zero

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Solution

The correct option is A $4 \times 10^{6}$
$v = \sqrt{\frac{2 e v}{m}}$
$= \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 45.5}{9.1 \times 10^{- 32}}} = 4 \times 10^{6} m / s$

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An electron is accelerated through potential difference of 45.5 volt. The velocity acquired by it is (in ms−1) :

The correct option is A 4×106v=√2evm=√2×1.6×10−19×45.59.1×10−32=4×106m/s

An electron is accelerated through potential difference of 45.5 volt. The velocity acquired by it is (in ${m s}^{- 1}$ ) :

The correct option is A $4 \times 10^{6}$
$v = \sqrt{\frac{2 e v}{m}}$
$= \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 45.5}{9.1 \times 10^{- 32}}} = 4 \times 10^{6} m / s$