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2.57    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\times 10^6 \ \textup{ms}^{-1}, calculate de Broglie wavelength associated with this electron.

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According to de- Broglie's equation for the wavelength.

\lambda = \frac{h}{mv}

Given the velocity of electron v = 1.6\times10^{6} m/s

and mass of electron m = 9.11\times10^{-31}kg

So, the wavelength will be:

\lambda = \frac{(6.626\times10^{-34}Js)}{(9.11\times10^{-31}kg)(1.6\times10^6 m/s)}

= 4.55\times10^{-10}m   or  455\ pm

Posted by

Divya Prakash Singh

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