Question

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

Answer 1

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 * $10^{-9}$ m

h = $6.626 * 10^{-34}$  J s

e = 1.6 * $10^{-19}$ C

m = 9.1 * $10^{-31}$ kg

Energy = eV

lambda = h / $\sqrt{2mE}$ = h / $\sqrt{2m(eV)}$

$(lambda)^{2}$ = $h^{2}$ / (2m (eV))

V = $h^{2}$ / (2 m e  $(lambda)^{2}$ )

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

V = 0.063 V

To learn more about wavelength of an electron  here

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