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Calculate the de Broglie wavelength of ...

Calculate the de Broglie wavelength of an electron moving with a speed that is 1% of the speed of light.

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`lamda=(h)/(mv)=(6.626xx10^(-34))/((9.11xx10^(-31))xx(3xx10^(8)xx(1)/(100)))`
`=2.42xx10^(-10)m`
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The french physicist Louis de Broglie in 1924 postulated that matter like radiation show a dual behaviour . He proposed the following relationship between the wavelength lamda of a material particle its linear momentum p and planck constant h lamda=h/p =h/(mv) The de broglie relation implies that the wavelength of a particle should decreases as its velocity increases . it also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de broglie waves. These waves differ from the electromagnetic waves as they (i) have lower velocities (ii) have no electrical and magnetic fields and (iii) are not emitted by the particle under consideration . The experimental confirmation of the de-broglie relation was obtained when Davisson ans Germer in 1927 observed that a beam of electrons is diffracted by a nickel crystal . as diffraction a characteristics property of waves hence the beam of electron behaves as a wave, as proposed by de-broglie. de- Broglie wavelength of an electron travelling with speed equal to 1% of the speed of light

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