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 $2 MeV$
As we know that, $V = \frac{W}{q}$
Charge on the alpha particle is $q = 2 e$

Kinetic energy $=$ work done $= q V = 2 e V = 2 \times 10^{6} e V = 2 MeV$

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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 MeVAs we know that, V=Wq Charge on the alpha particle is q=2e Kinetic energy = work done =qV=2eV=2×106e V=2 MeV

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

The correct option is B $2 MeV$
As we know that, $V = \frac{W}{q}$
Charge on the alpha particle is $q = 2 e$

Kinetic energy $=$ work done $= q V = 2 e V = 2 \times 10^{6} e V = 2 MeV$