Question

Li and a proton are accelerated by the same potential then their de Broglie wavelengths ${\lambda }_{Li}and{\lambda }_p$ have the ratio (assume $m_{Li}=9_{m_p})$

• A. $1:2$
• B. $1:4$
• C. $1:1$
• D. $1:3\sqrt{3}$

Answer 1

The correct option is D

$1:3\sqrt{3}$

${\lambda }_p=\frac{h}{\sqrt{2eV{m}_p}};{\lambda }_{Li}=\frac{h}{\sqrt{2\times 3eV{m}_{Li}}}=\frac{h}{\sqrt{2\times 3eV\times 9m_p}}Hence,\frac{{\lambda }_{Li}^{3+}}{{\lambda }_p}=\sqrt{\frac{2eV{m}_p}{2\times 3eV\times 9m_p}}=\frac{1}{3\sqrt{3}}$