Modern Physics Ques 154
- 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 $\frac{\lambda _1}{\lambda _2}$ is proportional to $(2004,2$
(a) $E^{0}$
(b) $E^{1 / 2}$
(c) $E^{-1}$
(d) $E^{-2}$
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Answer:
Correct Answer: 154.(b)
Solution:
Formula:
- $\frac{\lambda _1}{\lambda _2}=\frac{\frac{h}{\sqrt{2 m E}}}{\frac{h c}{E}}\quad $ or $\quad \frac{\lambda _1}{\lambda _2} \propto E^{1 / 2}$
BETA
