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

1 M e V

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2 M e V

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4 M e V

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8 M e V

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Solution

The correct option is B $2 M e V$
Charge on $α$ particle is $q = 2 e$ .
Kinetic energy $=$ work done $= q V = 2 e V = 2 \times 10^{6} e V = 2 M e V$

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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 MeVCharge on α particle is q=2e.Kinetic energy = work done =qV=2eV=2×106eV=2MeV

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

The correct option is B $2 M e V$
Charge on $α$ particle is $q = 2 e$ .
Kinetic energy $=$ work done $= q V = 2 e V = 2 \times 10^{6} e V = 2 M e V$