Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity v is given by
λ=hmv (∵p=mv)
de-Broglie wavelength of a proton of mass m1 and kinetic energy k is given by
λ1=h√2m1k (∵p=√2mk)
λ1=h√2m1qV....(i) [∵k=qV]
For an alpha particle mass m2 carrying charge q0 is accelerated through potential V, then
λ2=h√2m2q0V
∵ For α−particle (42He) : q0=2q and m2=4m1
∴λ2=h√2×4m1×2q×V....(ii)
The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get
λ1λ2=h√2m1qV×√2×m1×4×2qVh=4√2×√2√2
We get λ1λ2=2√2