Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100 \ V$ is

According to the de-Broglie hypothesis,the wavelength associated with a moving electron of mass $m$ is $\lambda_e$. Using the mass-energy relation and Planck's quantum theory,the wavelength associated with a photon is $\lambda_p$. If the energy $(E)$ of the electron and the photon is the same,then the relation between $\lambda_e$ and $\lambda_p$ is:

Photons of wavelength $\lambda$ are incident on the cathode of a photocell. Electrons are emitted from the cathode surface. The de-Broglie wavelength of the emitted electrons is (work function is negligible).
($c =$ velocity of light,$h =$ Planck's constant,$m =$ mass of electron)

The de-Broglie wavelength of a body of mass $m$ and kinetic energy $E$ is given by:

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 $pm$)