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

$A$ light of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. The photoelectrons emitted from the surface have de-Broglie wavelength $\lambda_1$. Then the ratio $\lambda : \lambda_1^2$ is ($h =$ Planck's constant,$c =$ velocity of light,$m =$ mass of electron).

If the mass of a neutron is $1.7 \times 10^{-27} \; kg$,then the de-Broglie wavelength of a neutron with energy $3 \; eV$ is (given $h = 6.6 \times 10^{-34} \; J \cdot s$):

$A$ particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$. Calculate the particle's mass and identify the particle.

The de Broglie wavelength for a deuteron can be given by:

Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will