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  • 29 The energy of a photon is equal to the kinetic energy of a proton. The energy of the proton is E. Let \lambda, be the de-Broglic wave length of the proton and <img alt="\lambda_2" src="https://e

29 The energy of a photon is equal to the kinetic energy of a proton. The energy of the proton is E. Let \lambda, be the de-Broglic wave length of the proton and \lambda_2 be the wave length of the photon. The ratio \frac{\lambda_1}{\lambda_2} is proportional to -
 

Option: 1

\mathrm{E^{\circ}}


Option: 2

\mathrm{E^{1/2} }


Option: 3

\mathrm{E^{-1} }


Option: 4

\mathrm{E^{-2}}


Answers (1)

best_answer

\begin{aligned} &\mathrm{Use, \lambda_1=\frac{h}{\sqrt{2 m E}}, \lambda_2=\frac{h c}{E}}\\ &\mathrm{ \frac{\lambda_1}{\lambda_2}=\frac{h}{\sqrt{2 m E}} \times \frac{E}{h c} }\\ &\mathrm{ \frac{\partial_1}{\lambda_2}=\frac{1}{c} \sqrt{\frac{E}{2 m}} }\\ &\mathrm{ \frac{\lambda_1}{\partial_2} \alpha \sqrt{E} \propto E Y_2 \text { Ans. }}\\ \end{aligned}

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