Maximum velocity of photo electrons emitted by a photo meter is 1.8 x 106 m/s.  Taking  em=1.8×1011 C/kg for electrons, the stopping potential of emitter is _____

Maximum velocity of photo electrons emitted by a photo meter is 1.8 x 106 m/s.  Taking  em=1.8×1011 C/kg for electrons, the stopping potential of emitter is _____

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    Solution:

    12mv2=e.V=v22em=(1.8)2×10122×1.8×1011=9 volt 

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