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Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 10^{6} ms^{–1}, calculate de Broglie wavelength associated with this electron.

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#### Solution 1

From de Broglie’s equation,

`lambda = "h"/("mv")`

`lambda = (6.626 xx 10^(-34) "Js")/(9.10939 xx 10^(-31)(1.6xx10^6 " ms"^(-1)))`

= 4.55 × 10^{–10} m

λ = 455 pm

∴ de Broglie’s wavelength associated with the electron is 455 pm.

#### Solution 2

λ = h/mv = 6.626×10^{-34 }kgm^{2}s^{-1}/(9.11×10^{-31} kg) (1.6×10^{6 }ms^{-1}) = 4.55×10^{-10 }m = 455 pm

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