According to de Broglie the wavelength of matter waves is given by

lambda = h/(mnu) = h/p`

Here h is Planck's constant , m is the mass , `nu` is the velocity of the particle , p is the momentum of the particle.

Lets derive the de Broglie wavelength of an electron using the above formula.

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de Broglie wavelength of an electron - derivation

Whenan electron of mass m and charge e is accelerated through a potential difference V, then the energy eV is equal to kinetic energy if the electron.

1/2 mnu^2 = eV `

or `rArr nu = sqrt( (2eV)/m)`--------------------------> ( 1)

The de Broglie wavelength is `lambda = h/(m nu)`

Substituting the value of `nu`

lambda = h/(m sqrt( (2eV)/m))

rArr lambda = h/(sqrt(2meV))` ------------------------> (2)

Substituting the known values in equation (2) , we get

lambda = 12.27/sqrt(V)` A^{0}

If V = 100 voltrs, then `lambda ` = 1.227 A^{o} , the wavelength associated with an electron accelerated by 100 volts is 1.227 A^{o} .

Since E =eV is kinetic energy associated with the electron, the equation (2) becomes

rArr lambda = h/sqrt(2mE)`

de Broglie wavelength of an electron - problems

1) What is tjhe de Broglie wavelength of an electron , if the speed is 105 ms^{-1}. (Given m = 9.1 x 10^{-31} kg ; v = 105 ms^{-1} ; h = 6.626 x 10^{-34} Js)

**Solution : **

Given:Mass of the electron is m = 9.1 x 10^{-31} kg ;

Velocity of the electron is `nu` = 105 ms^{-1} ;

Planck's constant h = 6.626 x 10^{-34} Js

Formula `lambda = h/(m nu)`

Plugging in the values we get `lambda = (6.626 xx 10 ^(-34))/(9.1 xx 10^(-31) xx 10^5)`

rArr lambda = 72.81 A^0`

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Pro : 2 What is the de Broglie wave length of an electron of kinetic energy 120 eV?

Solution : Given Kinetic e energy = 120 eV = 120 x 1.6 x 10-19 J.

Also for an electron we know that

m = 9.1 x 10^{-31} kg ;

h = 6.626 x 10^{-34} Js

Formula `lambda = h/sqrt(2mE)`

Plugging in the known values we get,

rArr lambda = (6.626 xx 10^(-34) )/( sqrt ( 2 xx 9.1 xx 10^(-31) xx 120 xx 1.6 xx 10^(-19)))`

rArr lambda="1.121" xx 10^(-10) m`