The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ 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: ............... $\mathring{A}$

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ 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....... $KeV$.

An $\alpha$-particle,a proton,and an electron have the same kinetic energy. Which one of the following is correct in the case of their De-Broglie wavelength?

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 $/8$)