A proton and a $L{i}^{3+}$ nucleus are accelerated by the same potential. If  ${\lambda }_{Li}and{\lambda }_P$ denote the de-Broglie wavelengths of  $L{i}^{3+}$and proton respectively, then the value of  $\frac{{\lambda }_{Li}}{{\lambda }_p}isx\times {10}^{-1}.$The value of x is…….. (Rounded off to the nearest integer) (Mass of $L{i}^{3+}=8.3$ mass of proton)

Give, mass of $L{i}^{3+}=8.3$ times of mass of proton formula, De-Broglie wavelength, 

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

Here, $h=$ Planck’s constant 

$=6.624\times {10}^{-34}J-s$

$m=$Mass of atom 

$q=$ Charge (or number of electrons)

${\lambda }_{Li}=\frac{h}{\sqrt{2m_{Li}\left(3e\right)V}}\mathrm{....}\left(i\right)$

${\lambda }_p=\frac{h}{\sqrt{2m_{Li}\left(e\right)V}}\mathrm{....}\left(ii\right)$

Now, Eq (i) divided by Eq. (ii), we get, $\frac{{\lambda }_{Li}}{{\lambda }_P}=\sqrt{\frac{m_p\left(e\right)V}{m_{Li}\left(3e\right)V}}$

We know that $m_{Li}=8.3m_p$

$\frac{{\lambda }_{Li}}{{\lambda }_p}=\sqrt{\frac{m_p\times e\times V}{8.3m_p\times 3e\times V}}$

$=\sqrt{\frac{1}{8.3\times 3}}=\frac{1}{5}=0.2=2\times {10}^{-1}$

Hence, $x\times {10}^{-1},x=2$