The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is \( E \). Let \( \lambda_1 \) be the de-Broglie wavelength of the proton and \( \lambda_2 \) be the wavelength of the photon. The ratio \( \lambda_1/\lambda_2 \) is proportional to the following
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( E^{1/2} \)
The de-Broglie wavelength of a particle of momentum \( p \) is \( \lambda = h/p \), and for a proton of kinetic energy \( E \) the momentum is \( p = \sqrt{2mE} \). A photon of energy \( E \) has wavelength \( \lambda_2 = hc/E \).
\( \displaystyle \lambda_1 = \frac{h}{\sqrt{2mE}} \qquad \lambda_2 = \frac{hc}{E} \)
\( \displaystyle \frac{\lambda_1}{\lambda_2} = \frac{h/\sqrt{2mE}}{hc/E} = \frac{\sqrt{E}}{c\sqrt{2m}} \)
Since \( m \) and \( c \) are constants, \( \lambda_1/\lambda_2 \propto E^{1/2} \).