A proton and a Li³⁺ nucleus are accelerated by the same potential. If $\lambda _{Li}$ and $\lambda _p$ denote the de Broglie wavelengths of Li³⁺ and proton respectively, then the value of ${{{\lambda _{Li}}} \over {{\lambda _p}}}$ is x $\times$ 10^-1.
The value of x is ______. (Rounded off to the nearest integer)
[Mass of Li³⁺ = 8.3 mass of proton]

Given, mass of Li³⁺ = 8.3 times of mass of proton formula,<br/><br/>De-Broglie wavelength, $\lambda = {h \over {\sqrt {2mqV} }}$<br/><br/>Here, h = Planck's constant = 6.624 $\times$ 10^$-$34 J-s<br/><br/>m = Mass of atom<br/><br/>q = Charge (or number of electrons)<br/><br/>${\lambda _{Li}} = {h \over {\sqrt {2{m_{Li}}(3e)v} }}$ .... (i)<br/><br/>${\lambda _P} = {h \over {\sqrt {2{m_P}(e)v} }}$ .... (ii)<br/><br/>Now, Eq. (i) divided by Eq. (ii), we get,<br/><br/>$${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P}(e)v} \over {{m_{Li}}(3e)v}}} $$<br/><br/>We know that m_Li = 8.3 m_p<br/><br/>$${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P} \times e \times V} \over {8.3\,{m_P} \times 3e \times V}}} = \sqrt {{1 \over {8.3 \times 3}}} = {1 \over 5} = 0.2$$ = 2 $\times$ 10^$-$1<br/><br/>Hence, x $\times$ 10^$-$1, x = 2