Calculate the de-Broglie's wavelength associated with an electron of energy 200eV .What will be the change in its wavelength if the accelerating potential is increased to four times its earlier value?
de - Broghe's wavelength is given by
Where h = Plank constant
P = momentum
P = mv
P2 = (mv)2 =
Given that KE = 200 eV
me = 9.11 × 10–31 kg
e = 1.6 × 10–19 C
P = 7.63 × 10–24 kg-m/s.
Hence, wavelength of the electron
It the accelerating voltage of electrons is increased to four times then its K.E. also increases to four times.
Hence momentum is increased twice.
P' = 2P
So new wavelength of the electron