Suppose you need to image the structure of a virus with a diameter of 50 nm. For
ID: 1393590 • Letter: S
Question
Suppose you need to image the structure of a virus with a diameter of 50 nm. For a sharp image, the wavelength of the probing wave must be 5.0 nm or less. We have seen that, for imaging such small objects, this short wavelength is obtained by using an electron beam in an electron microscope. Why don't we simply use short-wavelength electromagnetic waves? There's a problem with this approach: As the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. Let's compare the energy of a photon and an electron that can provide the same resolution.
For an electron with a de Broglie wavelength of 4.5nm , what is the kinetic energy (in eV)?
Express your answer using two significant figures.
Explanation / Answer
Kinetic energy of an electron
E = hc/lambda ......(1)
hc = 1240 eV nm
lambda = 4.5nm
put all these value in equation (1)
E = 1240 eV nm/4.5nm = 275.56 eV
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