Asked by David E
What potential difference is required in an electron microscope to give electrons a wavelength of 0.05 nm ?
Answers
Answered by
drwls
For that de Broglie wavelength, Electron momentum = p = h/(5*10^-11 m)
where h is Planck's constant.
Momentum p = 6.62*10^-34/5*10^-11
= 1.32*10^-23 kg*m/s
Energy = p^2/(2m) = e*V
m = electron mass
e = electron charge
Solve for potential difference, V.
V = p^2/(2*m*e)
where h is Planck's constant.
Momentum p = 6.62*10^-34/5*10^-11
= 1.32*10^-23 kg*m/s
Energy = p^2/(2m) = e*V
m = electron mass
e = electron charge
Solve for potential difference, V.
V = p^2/(2*m*e)
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