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{${2{m_0}{c^2}}$}}} }}

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