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 another type of material. If the velocity of the electron in this microscope is $$1.6 \times 10^{6}ms^{-1}$$, calculate de Broglie wavelength associated with this electron.

Asked by Abhisek | 1 year ago |  75

##### Solution :-

As per de Broglie’s equation,

$$\lambda =\dfrac{h}{m\nu }\\ \\ =\dfrac{(6.626\times 10^{-34})}{9.103939\times 10^{-31}kg(1.6\times 10^{6}ms^{-1})}\\ \\ =4.55\times 10^{-10}m\lambda =455pm$$

Therefore, de Broglie wavelength of the electron = 455 pm.

Answered by Abhisek | 1 year ago

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