28.56 Suppose you need to image the structure of a virus with a diameter of 50 n

Question

28.56 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.

Part A

For light of wavelength 5.0 nm , what is the energy (in eV) of a single photon? In what part of the electromagnetic spectrum is this?

E=__eV

Part B

For an electron with a de Broglie wavelength of 5.0 nm , what is the kinetic energy (in eV)?

E=__eV

Explanation / Answer

A photon is characterized by either a wavelength, denoted by or equivalently an energy, denoted by E. There is an inverse relationship between the energy of a photon (E) and the wavelength of the light () given by the equation:

E=hc/

where h is Planck's constant and c is the speed of light

h = 6.626 × 10 -34 joule·s

c = 2.998 × 108 m/s

By multiplying to get a single expression, hc = 1.99 × 10-25 joules-m

Therefore, we can rewrite the above constant for hc in terms of eV:

hc = (1.99 × 10-25 joules-m) × (1ev/1.602 × 10-19 joules) = 1.24 × 10-6  eV-m

Hence as per our data we have wavelength of 5nm

and Energy =  1.24 × 10-6  eV-m / 5 * 10^-9 = 248 eV (approx)

That puts it in the soft X-rays of SX class of the eletromagnetic spectrum.

Secondly

Kinetic energy of an electron having 4.5 nm wavelength, measured in electron volts is
E = hc/ =  248 eV

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