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A photon and a proton have the same de Broglie wavelength. Show, by actual calculations, which has more total energy.

 

 
 
 
 
 

Answers (1)

The energy of the photon

 E_{1}=\frac{hc}{\lambda }

the de-Broglie wave length of the proton ,

  \lambda =\frac{h}{m_{p}V}\Rightarrow m_{p}=\frac{h}{\lambda V}

The energy of the proton,

                    E_{2}=m_{p}C^{2}=\frac{h}{\lambda V}C^{2}

                    \frac{E_{2}}{E_{1}}=\frac{h}{\lambda V}C^{2}\times \frac{\lambda }{hc}

                           =\frac{C}{V}

                      E_{2}=\frac{C}{V}E_{1}

                        E_{2}> E_{1}

Posted by

Safeer PP

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