Please show work, no one else has got this right that I can find. Seriously, not
ID: 1517289 • Letter: P
Question
Please show work, no one else has got this right that I can find. Seriously, not one other tutor on chegg has got this right, according to webassign. Every formula I have tried has also been wrong.
Electrons in an electron microscope have a kinetic energy of 3.50*10^5 eV.
(a) Find the de Broglie wavelength of the electrons.
(b) Find the ratio of this wavelength to the wavelength of light at the middle of the visible spectrum (550 nm).
(c) How many times greater magnification is theoretically possible with this microscope than with a light microscope?
Explanation / Answer
E = 3.5*10^5*1.6*10^-19 = 5.6*10^-14 J
deBroglies wavelength is lamda = h/(sqrt(2*m*E) = (6.626*10^-34)/sqrt(2*9.11*10^-31*5.6*10^-14)
lamda = 2.07*10^-12 m = 2.07 pm
b) required ratio is (2.07*10^-12)/(550*10^-9) = 3.76*10^-6
C) magnification is 1/(3.76*10^-6) = 265957