તેથી \({\rm{K}}{\rm{.E}}{{\rm{.}}_{\rm{1}}} = qV = K.E{._2} = \frac{{p_1^2}}{{2{m_1}}} = \frac{{p_2^2}}{{2{m_2}}}\)
\( \Rightarrow \,\frac{{p_1^2}}{{p_2^2}} = \frac{{{m_1}}}{{{m_2}}}\,\,\,\,\,\,\frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \,\)
દ બ્રોગ્લી તરંગ લંબાઇ \(\, \Rightarrow \,\,\,\lambda \,\, = \,\,h/p\)
\(\frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{{p_2}}}{{{p_1}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} \)