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  • The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is . Let <

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is \mathrm{E}. Let \mathrm{\lambda _1} be the de-Broglie wavelength of the proton and \mathrm{\lambda _2}  be the wavelength of the photon. The ratio\mathrm{\lambda _1/\lambda _2}  is proportional to

Option: 1

\mathrm{E^{0}}


Option: 2

\mathrm{E^{1/2}}


Option: 3

\mathrm{E^{-1}}


Option: 4

\mathrm{E^{-2}}


Answers (1)

best_answer

\mathrm{\lambda _1=\frac{h}{\sqrt{2mE}}}\: \mathrm{and}\: \mathrm{\lambda _2=\frac{hc}{E}}

 \Rightarrow \frac{\lambda _1}{\lambda_2 } \ \alpha \ E^{\frac{1}{2}}

Posted by

HARSH KANKARIA

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