The idea of matter waves was given by

The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0, & 0 \le x \le 1 \\ 0, & x > 1 \end{cases}$. Let $\lambda_1$ and $\lambda_2$ be the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2E_0$,find $(\lambda_1/\lambda_2)^2$.

In an electron microscope,which nature of the electron is used?

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda_0$. What will be its wavelength at $927^{\circ} C$?

The de-Broglie wavelength of an electron is the same as that of a $50 \ keV$ $X$-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \ MeV$ ).