The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^8\, m/s$ is equal to the wavelength of a photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3 \times 10^8\, m/s$). (in $/8$)

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 $m_e c^2=0.5 \ MeV$)

When the kinetic energy of an electron is increased,the wavelength of the associated wave will:

The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?

An electron microscope uses electrons accelerated by a voltage of $50\; kV$. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture,etc.) are taken to be roughly the same,how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

The wavelength of a de-Broglie wave is $2 \mu m$. Calculate its momentum. (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$)