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

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