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The energy of a photon is equal to the kinetic energy of a proton. the energy of a photon is E. Let \lambda _{1} be the de Broglie wavelength of the proton and  \lambda _{2}be the wavelength of the photon then \lambda _{1}/\lambda _{2} is proportional to

Option: 1

E^{\circ}

 

 

 


Option: 2

E^{1/2}


Option: 3

E^{-1}


Option: 4

E^{-2}


Answers (1)

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As we learned

 

De - Broglie wavelength -

\lambda\: \alpha \: \frac{1}{p}\: \alpha \frac{1}{v}\: \alpha \frac{1}{\sqrt{E}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

 

 

\lambda _{1}=\frac{h}{P}=\frac{h}{\sqrt{2mE}}

\lambda _{2}=\frac{hc}{E}

\Rightarrow \frac{\lambda _{1}}{\lambda _{2}}\alpha \frac{E}{\sqrt{E}}=E^{1/2}

Posted by

Ritika Kankaria

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