A proton and an α-particle are accelerated, using the same potential difference. How are the de Broglie wavelengths λp and λa related to each other?
Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity v is given by λ=mvh (∵p=mv) de-Broglie wavelength of a proton of mass m1 and kinetic energy k is given by λ1=2m1kh (∵p=2mk) λ1=2m1qVh....(i) [∵k=qV] For an alpha particle mass m2 carrying charge q0 is accelerated through potential V, then λ2=2m2q0Vh ∵ For α−particle (24He) : q0=2q and m2=4m1 ∴λ2=2×4m1×2q×Vh....(ii) The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get λ2λ1=2m1qVh×h2×m1×4×2qV=24×22 We get λ2λ1=22
