The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \text{ Å}$ is (in $\text{ V}$)

The de-Broglie wavelength of an electron (mass $= 1 \times 10^{-30} \ kg$, charge $= 1.6 \times 10^{-19} \ C$) with a kinetic energy of $200 \ eV$ is (Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$):

Neglecting the variation of mass with velocity,the wavelength associated with an electron having a kinetic energy $E$ is proportional to

The kinetic energy of an electron is $5 \ eV$. Calculate the de-Broglie wavelength associated with it in $\mathring{A}$. $(h = 6.6 \times 10^{-34} \ J \cdot s, m_e = 9.1 \times 10^{-31} \ kg)$

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 with speed $v$ and a photon with speed $c$ have the same $de-Broglie$ wavelength. If the kinetic energy and momentum of the electron are $E_{e}$ and $p_{e}$ and that of the photon are $E_{ph}$ and $p_{ph}$ respectively,which of the following is correct?