Question

Biological specimens are often examined by microscopy. However, microscopy with visible light is limited to viewing details of a specimen on the order of the wavelength of light (400 nm). The macromolecules that make up the cell are much smaller, often about 2-10 nm in size, and the interatomic bonds that make up the molecular structure are about 0.2 nm in length. One method that can be used to reveal the atomic-level details of biological molecules is electron microscopy, in which a beam of electrons is focused onto a biological sample. Modern electron microscopes can emit a beam of electrons with a velocity of 1.5×108m/s.

a. What is the wavelength of an electron particle in this beam? Is the wavelength short enough to reveal molecular details at the atomic level?

b. The wavelength of the particle determines the resolution of the microscopy that can be performed. Assume that you desire a minimum uncertainty in the position of the electron of 1.0 Ǻ, -10 m. Using the uncertainty principle, what is the max imumuncertainty that is accept able in the momentum of the particle?

Answer 1

(a) Wavelength = 4.85 x 10^-12 m

(b) Maximum uncertainty = 5.27*10^-25 kgm/s

Step-by-step explanation:

(a) Calculating the wavelength using the formula;

λ= h/p

but p = mv

Where;

m = mass of electron = 9.1 x 10-31

h = 6.625 x 10^-34

= h/mv

= 6.625 x 10^-34 / (9.1 x 10^-31 x 1.5×10^8)

= 4.85 x 10^-12 m

so, the wavelength is short enough to reveal molecular details at the atomic level.

(b) Find attached of part b