Question

The mass of $L{i}^{3+}$is $8.33$ times the mass of a proton. If $L{i}^{3+}$ and proton are accelerated through the same potential difference, then the ratio of de Broglie's wavelength $L{i}^{3+}$of to proton is$x\times {10}^{-1}$. Find $x$

Answer 1

Step 1 : Given data

${\mathrm{m}}_{\mathrm{Li}}=8.3{\mathrm{m}}_{\mathrm{P}}$

Step 2: Applying De-Broglie wavelength

The wavelength equation is

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

$\frac{{\lambda }_{\mathrm{Li}}}{{\lambda }_{\mathrm{P}}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{P}}\left(\mathrm{e}\right)\left(\mathrm{V}\right)}{{\mathrm{m}}_{\mathrm{Li}}\left(3\mathrm{e}\right)\left(\mathrm{V}\right)}}$

Substitute the given data values are

$\frac{{\lambda }_{\mathrm{Li}}}{{\lambda }_{\mathrm{P}}}=\sqrt{\frac{1}{8.3\times 3}}$

$\frac{{\lambda }_{\mathrm{Li}}}{{\lambda }_{\mathrm{P}}}=$$\frac{1}{5}$

$\frac{{\lambda }_{\mathrm{Li}}}{{\lambda }_{\mathrm{P}}}=$=0.2

$\frac{{\lambda }_{Li}}{{\lambda }_P}=$$=2\times {10}^{-1}$

$x=2$

Therefore the value for $x=2$