Find the de Broglie wavelength of an electron in a metal at $27^{\circ}C$.

The ratio of the de-Broglie wavelength of an $\alpha$-particle to that of a proton,when both are subjected to the same magnetic field such that the radii of their paths are equal,assuming the magnetic field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton,is:

Crystal diffraction experiments can be performed using $X-$rays,or electrons accelerated through an appropriate voltage. Which probe has greater energy? (For quantitative comparison,take the wavelength of the probe equal to $1\,\mathring{A}$,which is of the order of inter-atomic spacing in the lattice) $(m_{e}=9.11 \times 10^{-31}\,kg)$

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

$A$ ball of mass $0.12 \ kg$ is moving with a speed $20 \ m \ s^{-1}$. Then its de Broglie wavelength is . . . . . . . ( $h = 6.63 \times 10^{-34} \ J \ s$ )

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{e} / \lambda_{p}$ is