Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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
#SPJ4
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.