Question

Physical Chemistry

5. Because of their short wavelength, the resolution of an electron microscope can be made several 100-fold higher than that of a microscope based on light. They have thus become one of the most powerful techniques for studying macromolecules, assemblies of macromolecules, cellular organizations, and so on. Modern electron microscopes can emit a beam of electrons with a velocity of 1.5 x 10 m/sec. a) Enter the wavelength of an electron in this beam in meters and to two significant figures. Do not enter the units. Recall that p mv, the mass of the electron is 9.11 x 10 kg, and 1 Joule- 1 kg m2/s2. (Even though the wavelength is short enough to reveal atomic dimensions in practice this is not achieved.) b) Assume that you desire a uncertainty in the position of the electron of 1 10 10 m). Using the uncertainty principle, what is the maximum uncertainty that is acceptable in the velocity of the particle? What of the beam velocity above does this correspond to? Enter the value of the uncertainty in m/s and the percent in that order and two significant figures. Do not enter the units.

Answer 1

according to deBroglie equation $\lambda$ =h/mv

given m= 9.11 x10^-31 Kg

v = 1.5 x 10⁸ m/s

we know h= Plancks constant = 6.626 x 10^-34 J.s

substituting values we get $\lambda$ = (6.626 x 10^-34) J.s/ 9.11 x10^-31 Kg x 1.5 x 10⁸ m/s

= 0.484 x 10^-11 m

= 4.8 x 10^-12 m

b. $\Delta$ x. $\Delta$ v=h/4 $\pi$ m

=given    $\Delta$ x = 10^-10m, h= 6.626 x 10^-34 J.s, m= 9.11 x 10^-31 Kg

substituting the values , we get $\Delta$ v

$\Delta$ v=h/4 $\pi$ m $\Delta$ x

=  6.626 x 10^-34 J.s/ 4 x 3.14 x 9.11 x 10^-31 x 10^-10

= 0.0579 x 10⁷ m/s

% beam velocity =[ $\Delta$ v/ velocity of electron beam] x 100  

=  0.0579 x 10⁷ x 100/ 1.5 x 10⁸

= 0.386

=0.39%