Question

You would like to observe an E. coli bacterium that is 1.85 μm1.85 μm long. Because diffraction blurs an image, you would like to minimize the effects of diffraction by using a wavelength no larger than the object you are observing. For this problem, assume the wavelength is equal to the length of the bacterium.

What is the energy of a photon with this wavelength?

the energy of photon: ______ JJ

What is the energy of an electron with a de Broglie wavelength of this size?

the energy of electron: _____JJ

In terms of energy, which particles, photons, or electrons, are the least likely to damage your sensitive biological sample?

a. electrons

b. they are the same

c. photons

Answer 1

1.

Wavelength of photon, λ = 1.85 x 10^-6 m

Energy of photon, E = hc/λ = 6.625 x 10^-34 x 3 x 10⁸/1.85 x 10^-6 = 1.074 x 10^-19 J

Energy of photon, E = 1.074 x 10^-19 J

2.

De-Broglie wavelength of electron, λ = 1.85 x 10^-6 m

De-Broglie wavelength of electron, λ = h/(2mE)^1/2

Energy of electron, E = h²/2m λ² = (6.625 x 10^-34 )²/2x 9.11 x 10^-31 x (1.85 x 10^-6)² = 7.03x10^-26J

Energy of electron, E = 7.03x10^-26J

3.

Since, the energy of electron is much less than the energy of photon, the electron is least likely to damage sensitive biological sample.