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If an electron having kinetic energy 2 e...

If an electron having kinetic energy `2 eV` is accelerated through the potential difference of `2` volt. Then calculate the wavelength associated with the electron

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The correct Answer is:
`6.15 Å`
Initial kinetic energy `KE=2 ev`
increase in `KE` due to acceleration `=qxxV=exx2v=2eV`
`:.` Fimal kinetic energy, `KE_(f)=2+2=4 eV`.
`:.` Associated de-Broglie wavelength, `lambda=12.27/sqrt(4)=6.15 Å`
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