Calculate the de Broglie wavelength of an electron with a kinetic energy of $200 \ eV$. [Mass of electron $= 1 \times 10^{-30} \ kg$,charge on electron $= 1.6 \times 10^{-19} \ C$,Planck's constant $(h) = 6.6 \times 10^{-34} \ Js$].

$A$ particle $A$ of mass $m$ and charge $q$ is accelerated by a potential difference of $50 \ V$. Another particle $B$ of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500 \ V$. The ratio of de-Broglie wavelength $\frac{\lambda_A}{\lambda_B}$ is close to

The de-Broglie wavelength of a body of mass $1 \ kg$ moving with a velocity of $2000 \ m/s$ is:

The de-Broglie wavelength of a particle accelerated with $150 \ V$ potential is $10^{-10} \ m$. If it is accelerated by $600 \ V$ potential difference,its wavelength will be: ............... $\mathring{A}$

Two particles have the same charge. If they are accelerated through the same potential difference,what will be the ratio of their de Broglie wavelengths?

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?