Question

Nuclear radii may be measured by scattering high-energy electrons from nuclei. (a) What is the de Broglie wavelength for 287 MeV electrons? (b) Are these electrons suitable probes for this purpose?

Answer 1

Part A)

Here energy = 287 Mev = 4.59 * 10^-11 J can be taken as Kinetic Energy of electron.

Now De- broglie wavelenth = h/p here h is planks constant , and p is momentum

or wavelenth= h/ 2mK , Here m= mass and K= kinetic energy

so wavelength= (6.63 * 10^-34)   / ( 2 * 9.1 * 10^-31 * 4.59 * 10^-11 )

So Wavelength = 1.12 * 10 ^-14 m

Part B)

Yes ,the electron is a better nuclear probe than the other(alpha particles of Rutherford scattering) because it is a point particle and can penetrate the nucleus.