de
Broglie wavelength for an electron
λe=
h / √(2mE) ;
λe ∝ 1/√E ;
λe2 ∝ 1/E ------------- (1)
For
photon
λp=
hc / E ;
λp ∝ 1/E ------------- (2)
From
equation (1) and (2), we have So,
λp ∝ λe2
When
alpha particle is accelerated at the potential V;
K.E
= 2eV