An alpha particle is accelerated through a potential difference of $10^{6}$ volt. Its kinetic energy will be

1 MeV

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2 MeV

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4 MeV

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8 MeV

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Solution

The correct option is B 2 MeV
$K = q V = 2 e \times 10^{6} = 2 M e V$

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An alpha particle is accelerated through a potential difference of 106 volt. Its kinetic energy will be

The correct option is B 2 MeVK=qV=2e×106=2MeV

An alpha particle is accelerated through a potential difference of $10^{6}$ volt. Its kinetic energy will be

The correct option is B 2 MeV
$K = q V = 2 e \times 10^{6} = 2 M e V$