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?

Hello,

de - Broghe's wavelength is given by

Where *h* = Plank constant

P = momentum

P = mv

P^{2} = (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.

Since

Hence momentum is increased twice.

P' = 2P

So new wavelength of the electron

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