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What potential difference is required in an electron microscope to give an electron wavelength of 4. 5 nm?

Sagot :

Potential difference required in an electron microscope to give an electron wavelength of 4. 5 nm will be 0.063 V.

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other is called potential difference.

The wavelength of an electron is calculated for a given energy (accelerating voltage) by using the de Broglie relation between the momentum p and the wavelength λ of an electron

lambda = 4.5 nm = 4.5 * [tex]10^{-9}[/tex] m

h = [tex]6.626 * 10^{-34}[/tex]  J s

e = 1.6 * [tex]10^{-19}[/tex] C

m = 9.1 * [tex]10^{-31}[/tex] kg

Energy = eV

lambda = h / [tex]\sqrt{2mE}[/tex] = h / [tex]\sqrt{2m(eV)}[/tex]

[tex](lambda)^{2}[/tex] = [tex]h^{2}[/tex] / (2m (eV))

V = [tex]h^{2}[/tex] / (2 m e  [tex](lambda)^{2}[/tex] )

V  =  [tex](6.626 * 10^{-34} )^{2}[/tex] /  2 * 9.1 * [tex]10^{-31}[/tex] *  1.6 * [tex]10^{-19}[/tex]  * [tex](4.9 * 10^{-9}) ^{2}[/tex]

V = 0.063 V

To learn more about wavelength of an electron  here

https://brainly.com/question/17295250

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