Question

in one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is

Answer 1

de-Broglie wavelength = h/$\sqrt (2meV)$

For proton e = 1eV that is accelerated by 3V

For alpha particle e=20eV that is retarded by 2V

For a proton, let mass = m

For alpha particle mass = 4m

Ratio of their wavelengths will be [h/$\sqrt (2m*1*3)$] / [h/$\sqrt (2*4m*20*4)$]

Ratio = 320 : 3