a7|when an electron of charge e is accelerated through a potential of V volts, the associated waveleng electron will be:- h -] (2meV) Alt1 h Alt2 (meV) 2met Alt3 7. = (2meV) 2. = h
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de-Broglie wavelength is given by λ=ph
where, momentum of the particle p=2mK
Kinetic energy of the particle K=eV
⟹ λ=2meVh
We get λ∝V1
Thus de-Broglie wavelength of the electron will become half if the accelerating potential is increased by a factor of 4.
I hope my answere is right, I'm not that sure tho
it's been a while since I read the chapter
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