Show that the de-Broglie wavelength `lambda` of electrons of energy E is givne by the relation:
` lambda= h /(sqrt(2 m E))`.

What is the de-Broglie wavelength of an electron with kinetic energy of 120 eV?

X-rays of wavelength lambda fall on an phtot senstive surface,emitting electrons.Assuming that the work function of the surface can be netglected,prove that the de-Broglie wavelength of electrons emited will be sqrt(h lambda//(2 m c)) .

If a_(0) be the radius of first Bohr's orbit of H-atom, the de-Broglie's wavelength of an electron revolving in the second Bohr's orbit will be: