Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get immediate and reliable solutions to your questions from a community of experienced professionals on our 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
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
