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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