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(B) For a proton,the de Broglie wavelength is given by $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For a photon,the energy is given by $E = pc$,s

Vedclass · Accepted Answer

$E^{1/2}$

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(B) For a proton,the de Broglie wavelength is given by $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For a photon,the energy is given by $E = pc$,so the momentum is $p = \frac{E}{c}$. The de Broglie wavelength is $\lambda_2 = \frac{h}{p} = \frac{hc}{E}$.Taking the ratio of the two wavelengths:$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} 	imes \frac{E}{hc} = \frac{1}{c\sqrt{2m}} 	imes \frac{E}{E^{1/2}} = \frac{1}{c\sqrt{2m}} 	imes E^{1/2}$.Thus,$\frac{\lambda_1}{\lambda_2} \propto E^{1/2}$.Therefore,the correct option is $B$.

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to:

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