The energy that should be added to an electron to reduce its de Broglie wavelength from $1 \, nm$ to $0.5 \, nm$ is

An electron beam of energy $10 \, KeV$ is incident on a metallic foil. If the interatomic distance is $0.2454 \, \mathring{A}$,then find the angle of diffraction in degrees. $(\sqrt{150} = 12.27)$

$A$ stationary mass $M$ splits into two fragments of masses $m_1$ and $m_2$. What is the ratio of their de Broglie wavelengths,${\lambda _1}/{\lambda _2}$?

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelength of the electron to the wavelength of the photon is ($c =$ velocity of light).

An electron of charge $e$ and mass $m$ moving with an initial velocity $v_0 \hat{i}$ is subjected to an electric field $E_0 \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron $= \lambda_0$)

An electron of mass $m$ is accelerated through a potential difference of $V$. Its de Broglie wavelength is $\lambda$. If a proton of mass $M$ is accelerated through the same potential difference,its de Broglie wavelength will be ..............