The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons would one need in an electron microscope to resolve separations of

(a) 150 Angstrom

(b) 5 Angstrom

Question

Physics

Mckenzie

30 August, 11:40

The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons would one need in an electron microscope to resolve separations of

(a) 150 Angstrom

(b) 5 Angstrom

+4

Answers (1)

Know the Answer?

Mike Mckenzie · Answer 1

Answer: a) for 150 Angstroms 6.63 * 10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ = p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

Finally we have:

eΔV=p^2/2m = h^2 / (2*m*λ^2)

replacing we obtained the above values.

The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons would one need in an electron microscope to resolve separations of(a) 150 Angstrom(b) 5 Angstrom

The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons would one need in an electron microscope to resolve separations of

(a) 150 Angstrom

(b) 5 Angstrom