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
Related Physics A Level answers
All answers ▸How would you calculate the moment of a Force on a rigid object?
Two balls of mass 3kg and 7 kg respectively move towards one another with speeds 5ms^-1 and 2ms^-1 respectively on a smooth table. If they collide and join, what velocity do they move off with?
Calculate the threshold wavelength for a metal surface with work function of 6.2 eV.
A ball is dropped from rest from a window 3m above ground height. How long will it take the ball to hit the ground? (You may assume air resistance on the ball is negligible.)