To resolve an object in an electron microscope, the wavelength of the electrons must be close to the diameter of the object. What kinetic energy, ?k�k , must the electrons have in order to resolve a protein molecule that is 4.30 nm4.30 nm in diameter? Take the mass of an electron to be 9.11×10−31 kg9.11×10−31 kg .

I need some help with the following problem. 1a. What is the maximum number of electrons that can be emitted if a potassium surface of work function 2.40 eV absorbs 5.00 x 10-3 J of radiation at a wavelength of 325 nm? b. What is the kinetic energy and velocity of the electrons emitted?

To resolve an object in an electron microscope, the wavelength of the electrons must be close to the diameter of the object. What kinetic energy must the electrons have in order to resolve a protein molecule that is 7.00 nm in diameter? Take the mass of an electron to be kg.

Cesium has a work function of 2.14 eV where 1.0 eV = 1.602 x 10^-9J. If radiation with a wavelength of 245 nm shines on a cesium surface what will be the de Broglie wavelength of the emitted electron?   a. 718 pm b. 728 pm c 710 pm d. 698 pm e. 678 pm

The de Broglie wavelength of an electron has to do with spatial resolution of an electron microscope, which is often expressed in the unit of length Å (Angstrom). The 1 V potential difference causes an electron to gain kinetic energy EK of 1 electron Volt (eV). In the SI units, kinetic energy in eV must be converted to Joules. The conversion factor is 1 eV = 1.6 x 10-19 Joule. And, the formula for the wavelength is: λ = h / p = h / √(2 m EK) where m is electron mass. Calculate the de Broglie wavelength of an electron (in Å) when the electron is accelerated from rest through a potential difference of: a) 1 kV = 1,000 V (a low resolution setting of microscope), b) 10 kV = 10,000 V (intermediate resolution). c) 100 kV = 100,000 V (high resolution),

An electron is located on a pinpoint having a diameter of 6.11 µm. What is the minimum uncertainty in the speed of the electron?

To resolve an object in an electron microscope, the electrons' wavelength must be close to the diameter of the object. What kinetic energy must the electrons have in order to resolve a protein molecule that is 8.10 nm in diameter? Take the mass of an electron to be 9.11 ×× 1031 kg.

D 11.5

Calculate the velocity (in m/s) of an electron that has a wavelength of 8.76 µm.  m/s (b) Through what voltage (in V) must the electron be accelerated to have this velocity? (Assume the electron starts at rest.)  V

An electron microscope passes 1.00-pm-wavelength electrons through a circular aperture 2.00 μm in diameter.What is the angle between two just-resolvable point sources for this microscope?