# Wavelength of a particle moving with relativistic speed

Question

Obtain expression for the wavelength of a particle moving with relativistic speed.

Solution

The relativistic momentum of a particle is given by

$p = \frac{{{m_0}v}}{{\sqrt {1 – {{{v^2}} \mathord{\left/ {\vphantom {{{v^2}} {{c^2}}}} \right. \kern-\nulldelimiterspace} {{c^2}}}} }}$

$\therefore \lambda = \frac{h}{p} = \frac{{h{{\left( {1 – {{{v^2}} \mathord{\left/ {\vphantom {{{v^2}} {{c^2}}}} \right. \kern-\nulldelimiterspace} {{c^2}}}} \right)}^{{\raise0.7ex\hbox{1} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{2}}}}}}{{{m_0}v}} = \frac{h}{{{m_0}c}}\frac{{{{\left( {1 – {{{v^2}} \mathord{\left/ {\vphantom {{{v^2}} {{c^2}}}} \right. \kern-\nulldelimiterspace} {{c^2}}}} \right)}^{{\raise0.7ex\hbox{1} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{2}}}}}}{{\left( {{\raise0.7ex\hbox{v} \!\mathord{\left/ {\vphantom {v c}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{c}}} \right)}}$

The momentum $p$ of a relativistic particle can also be expressed as follows

${E^2} = {p^2}{c^2} + {({m_0}{c^2})^2} = {(T + {m_0}{c^2})^2}$

$p = \frac{{\sqrt {T(T + {m_0}{c^2})} }}{c}$

Hence

$\lambda = \frac{h}{p} = \frac{{hc}}{{\sqrt {T(T + {m_0}{c^2})} }}$

$\lambda = \frac{h}{{\sqrt {2{m_0}T} }}\frac{1}{{\sqrt {1 + {\raise0.7ex\hbox{T} \!\mathord{\left/ {\vphantom {T {2{m_0}{c^2}}}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{{2{m_0}{c^2}}}}} }}$