Search any question & find its solution
Question:
Answered & Verified by Expert
If we assume kinetic energy of a proton is equal to energy of the photon, the ratio of de Broglie wavelength of proton to photon is proportional to
Options:
Solution:
1251 Upvotes
Verified Answer
The correct answer is:
$E^{1 / 2}$
For photon, $\lambda_P=\frac{h c}{E}$
Here, $E$ is energy of photon.
For proton, $\lambda_p=\frac{h}{\sqrt{2 m K}}$
Here, $m$ and $K$ be mass and kinetic energy of proton respectively.
$$
\therefore \quad \frac{\lambda_p}{\lambda_P}=\frac{E}{c} \sqrt{2 m K} \propto E^{1 / 2} \quad(\because E=K \text { Given })
$$
Here, $E$ is energy of photon.
For proton, $\lambda_p=\frac{h}{\sqrt{2 m K}}$
Here, $m$ and $K$ be mass and kinetic energy of proton respectively.
$$
\therefore \quad \frac{\lambda_p}{\lambda_P}=\frac{E}{c} \sqrt{2 m K} \propto E^{1 / 2} \quad(\because E=K \text { Given })
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.