TS EAMCET · Physics · Dual Nature of Matter
The energy of a photon is equal to the kinetic energy of a proton. If \(\lambda_1\) is the de-Broglie wavelength of a proton, \(\lambda_2\) the wavelength associated with the proton and if the energy of the photon is \(E\), then \(\left(\lambda_1 / \lambda_2\right)\) is proportional to
- A \(E^4\)
- B \(E^{1 / 2}\)
- C \(E^2\)
- D \(E\)
Answer & Solution
Correct Answer
(B) \(E^{1 / 2}\)
Step-by-step Solution
Detailed explanation
We know that, de-Broglie wavelength of photon \[ E=h \mathrm{v} \] But we know that also \[ \begin{aligned} E & =\frac{1}{2} m v^2 \\ \therefore \quad 2 E & =\frac{P^2}{m} \\ \because \quad P & =m v \\ \frac{P^2}{m} & =m v^2 \\ P & =\sqrt{2 m E} \end{aligned} \] (for proton)…
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