If the momentum of an electron is changed by $P_m$ and the associated de Broglie wavelength changes by $0.50\ \%$, find the initial momentum of the electron. (in $P_m$)

The ratio of de Broglie wavelength of a deuteron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$ is $n:1$. The value of $n$ is . . . . . . .
(Assume mass of proton = mass of neutron)

An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few $eV$. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ($c =$ speed of light in vacuum).

The de-Broglie wavelength of an electron having kinetic energy $100 eV$ is, $[$ Use $h=4.14 \times 10^{-15} eVs$, mass of electron $= \frac{0.5 \times 10^6}{c^2} eV/c^2$, $1 pm = 10^{-12} m$ $]$ (in $pm$)

The de-Broglie wavelength of an electron with kinetic energy of $320 eV$ is (Take $h = 6.0 \times 10^{-34} \text{ SI unit}$, mass of electron $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, charge of an electron $e = 1.6 \times 10^{-19} \text{ C}$). (in $pm$)

If the kinetic energies of an electron, an alpha particle, and a proton having the same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$, and $\varepsilon_3$ respectively, then: