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Question

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

A
1 MeV
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B
2 MeV
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C
4 MeV
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D
8 MeV
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Solution

The correct option is B 2 MeV
Charge on α particle is q=2e.
Kinetic energy = work done =qV=2eV=2×106eV=2MeV

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