The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

An electron,a doubly ionized helium ion $(He^{++})$ and a proton have the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{e}, \lambda_{He^{++}}$ and $\lambda_{P}$ is:

$A$ 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 an electron having kinetic energy $100 eV$ is, $[$ Use $h=4.14 \times 10^{-15} eVs$, mass of electron $= \frac{0.5 \times 10^6}{c^2} eV/c^2$, $1 pm = 10^{-12} m$ $]$ (in $pm$)

If the velocity of a moving particle is reduced to half, what will be the percentage change in its de Broglie wavelength?

The ratio of de Broglie wavelength of a deuteron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$ is $n:1$. The value of $n$ is . . . . . . .
(Assume mass of proton = mass of neutron)