The E. coli bacterium is about 2.0 mu m long. Suppose you want to study it using

Question

The E. coli bacterium is about 2.0 mu m long. Suppose you want to study it using photons of that wavelength or electrons having that de Broglie wavelength. What is the energy of the photon and the energy of the electron? photon energy__________J electron energy_________J Which one would be better to use, the photon or the electron? Explain why. SmallCircle The electron energy is over a million times greater than that of the photon. Therefore, the electron would be much more likely to damage (or destroy) the bacterium in the process of studying it. Thus, it would be better to use the photon. SmallCircle The photon energy is over a million times smaller than that of the electron. Therefore, the photon would be much more likely to damage (or destroy) the bacterium in the process of studying it. Thus, it would be better to use the electron. SmallCircle The small difference between the the photon energy and the electron energy is negligible. Thus, either method can be employed. SmallCircle The photon energy is over a million times greater than that of the electron. Therefore, the photon would be much more likely to damage (or destroy) the bacterium in the process of studying it. Thus, it would be better to use the electron.

Explanation / Answer

Given wavelength () = 2*10-6 m

a) Energy of photon is given by Ep = h*c/ = 6.63*10-34*3*108/2*10-6  = 9.945*10-20 J

de Broglie wavelength and momentum of electron are related by

p = h/ = 6.63*10-34/2*10-6  = 3.315*10-28  Kg-m/s

So the kinetic energy of electron Ee  = p2/2*m = (3.315*10-28)2/(2*9.1*10-31) = 6.038*10-26  J

(b) The correct option would be the last one i,e The photon energy is over a million times greater than electron...

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