Electrons in an electron microscope have a kinetic energy of 2.04 105 eV. (a) Fi

Question

Electrons in an electron microscope have a kinetic energy of 2.04 105 eV.

(a) Find the de Broglie wavelength of the electrons. 2.72e-12 m Incorrect: Your answer is incorrect.

(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?

Please show your work so I can understand this. There is a lot of different and incorrect answers on here!

Explanation / Answer

(a) The energy of the electron and it's wavelength can be related by: E = hc/L ( h is plank's constant, c is velocity of light in vacuum, L is wavelength)

So, L = hc/E = { 6.63*10^(-34) × 3*10^8}/ 3.268*10^(-14)

In the previous step i have converted energy E from eV to joule. 1ev = 1.602*10^(-19) joule. So 2.04*10^5 eV = 3.268*10^(-14) joule.

Therefore L comes out to be 6.08*10^(-12) m

(b) 1 nm = 10^(-9) m.

So 550 nm = 550×10^(-9) m = 5.5×10^(-7) m.

So the ratio of wavelength = 6.08×10^(-12)/5.5×10(-7) = 1.105×10^(-5).

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