Question

li+3 and a proton are accelerated by the same potential, their broglie wave length li and p have the ratio

Answer 1

De-broglie's wavelength equation in term of electric potential is given by,

$\lambda=\frac{h}{\sqrt{2qVm}}$

where $\lambda$ denotes the wavelength, q is charge, V is electric potential and m is mass of particle.

case 1: mass of proton, m = 1u

charge on proton , q = 1e

so, wavelength of proton = h/√{2.1eV.1u}

case 2 : mass of Li³+ = 7u

charge on Li³+ = 3e

so, wavelength of Li³+ = h/√{2.3e.V.7u}

now, wavelength of Li³+/wavelength of proton = [h/√{2.3e.V.7u}]/[h/√{2.1e.V.1u}]

= √{2/42}

= 1/√21

hence, answer should be 1 : √21