Electrons are accelerated through a potential difference of 300 V. What is their final de Broglie wavelength?
The kinetic energy of the electron will be equal to the energy what the electric field gives to it.1/2mv2=VQFrom thisv=sqrt(2VQ/m)
De Broglie wavelength:lambda=h/p=h/(mv) We know that h is the planck constant, V=300V Q is the charge of an electron and m is the mass of it.
If we substitute these into the equations above we get lambda=710-11m
CB
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