The de-Broglie wavelength of an electron is $0.4 \times 10^{-10} \ m$ when its kinetic energy is $1.0 \ keV$. Its wavelength will be $1.0 \times 10^{-10} \ m$ when its kinetic energy is: (in $keV$)

The de Broglie wavelength and kinetic energy of a particle are $2000 \ \mathring{A}$ and $1 \ \text{eV}$ respectively. If its kinetic energy becomes $1 \ \text{MeV}$,then its de Broglie wavelength becomes $...... \ \mathring{A}$.

The de Broglie wavelength of an electron of kinetic energy $9 \ eV$ is (take $h=4 \times 10^{-15} \ eV \cdot s$,$c=3 \times 10^{10} \ cm/s$ and the mass $m_e$ of electron as $m_e c^2=0.5 \ MeV$)

If $E_p$ and $E_e$ represent the kinetic energy of a photon and an electron respectively. If the de-Broglie wavelength $\lambda_p$ of a photon is twice the de-Broglie wavelength $\lambda_e$ of an electron,then $E_e / E_p$ is (Speed of electron $= C/100$,where $C$ is the velocity of light).

For which of the following particles will it be most difficult to experimentally verify the de Broglie relationship?

The accelerating voltage of an electron gun is $50,000 \ V$. What will be the de Broglie wavelength of the electron in $\mathring{A}$?