Question

The ratio of potential applied to accelerate the $\alpha$- particle and proton in order to have the same de Broglie wavelength is

Answer 1

$\lambda =\frac{h}{\sqrt{2m_1{q}_1{v}_1}}$

$\lambda =\frac{h}{\sqrt{2m_2{q}_2{v}_2}}$

$\Rightarrow m_1{q}_1{v}_1=m_2{q}_2{v}_2$

$\Rightarrow \frac{v_1}{v_2}=\frac{m_2{q}_2}{m_1{q}_1}$

$=\frac{1\times 1}{4\times 2}$

$\frac{v_1}{v_2}=\frac{1}{8}$

We get, $\frac{v_1}{v_2}=\frac{1}{8}.$