$A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles,$\lambda_1 / \lambda_2$ is

If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal,then the ratio of velocities of the proton and the $\alpha$-particle will be

$A$ beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 \, Å$. The first maximum intensity occurs at $\theta = 60^{\circ}$. Then $E$ (in $eV$) is: (Planck constant $h = 6.64 \times 10^{-34} \, Js$, $1 \, eV = 1.6 \times 10^{-19} \, J$, electron mass $m = 9.1 \times 10^{-31} \, kg$)

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-rays is

The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is $........$

$A$ particle of mass $1 \times 10^{-30} \,kg$ and electric charge $1.6 \times 10^{-19} \,C$ has a de-Broglie wavelength of $660 \,nm$. The kinetic energy of this particle is (Planck's constant,$h = 6.6 \times 10^{-34} \,J \cdot s$)