The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

$A$ proton and an $\alpha$-particle are accelerated using the same potential difference. How are the de-Broglie wavelengths $\lambda_p$ and $\lambda_{\alpha}$ related to each other?

If the de Broglie wavelength of an electron is $5200 \ \mathring{A}$,what is its velocity?

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?

What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$?

An object of mass $200 \ g$ is moving with a speed of $5 \ m/hr$. The order of the de Broglie wavelength associated with it is ...... $(h = 6.626 \times 10^{-34} \ J \cdot s)$.