Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$,then the value of $n$ is

An electron (mass $m$) with an initial velocity $\overrightarrow{v} = v_{0} \hat{i} \left(v_{0} > 0\right)$ is moving in an electric field $\overrightarrow{E} = -E_{0} \hat{i} \left(E_{0} > 0\right)$ where $E_{0}$ is constant. If at $t = 0$ the de Broglie wavelength is $\lambda_{0} = \frac{h}{mv_{0}}$,then its de Broglie wavelength after time $t$ is given by:

The ratio of de Broglie wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperatures of $27\,^oC$ and $127\,^oC$ respectively is

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ is:

What is the de-Broglie wavelength of a bullet of mass $0.033$ kg travelling at the speed of $1$ km/s? $(h = 6.6 \times 10^{-34} \text{ Js})$