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 of $L{i}^{3+}$ to proton is $x\times {10}^{-1}.$ Find x.

Answer 1

$m_{\left(L{i}^{3+}\right)}=8.33\times m_{p+}\left(given\right)$
Debroglie's wavelength (λ) = $\frac{h}{p}$
$K.E=\frac{1}{2}m{v}^2$
Multiplying by m on both sides
$2m\times K.E=(mv{)}^2=p^2$
$P=\sqrt{2m\times K.E}$
Also KE= q × V
So, $P=\sqrt{2m.q.V}$
$\lambda =\frac{h}{\sqrt{2mqV}}$
$\frac{{\lambda }_{L{i}^{3+}}}{{\lambda }_{p+}}=\frac{\frac{h}{\sqrt{2\times m_{L{i}^{3+}}\times 3\times V}}}{\frac{h}{\sqrt{2\times m_{p+}\times 1\times V}}}=\frac{\sqrt{m_{p+}}}{\sqrt{3\times m_{L{i}^{3+}}}}=\sqrt{\frac{m_{p+}}{3\times 8.33m_{p+}}}$
$\frac{{\lambda }_{L{i}^{3+}}}{{\lambda }_{p+}}=\frac{1}{\sqrt{25}}=\frac{1}{5}=0.2=2\times {10}^{-1}$
Comparing with x × ${10}^{-1}=2\times {10}^{-1}$
X = 2