$A$ particle is dropped from a height $H$. The de Broglie wavelength of the particle depends on height as

The de Broglie wavelengths associated with a proton and an electron are in the ratio $2: 1$. Their stopping potentials are approximately in the ratio of

The de-Broglie wavelength of a neutron at $27^oC$ is $\lambda$. What will be its wavelength at $927^oC$?

The de-Broglie wavelength of an electron (mass $= 1 \times 10^{-30} \ kg$, charge $= 1.6 \times 10^{-19} \ C$) with a kinetic energy of $200 \ eV$ is (Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$):

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

We wish to see inside an atom. Assuming the atom to have a diameter of $100 \ pm$, this means that one must be able to resolve a width of say $10 \ pm$. If an electron microscope is used, the minimum electron energy required is about....... $KeV$.