If the kinetic energy of a particle is increased to $16$ times its previous value,the percentage change in the de-Broglie wavelength of the particle is

The ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is:

$A$ proton moves in a circular path of radius $6.6 \times 10^{-3} \ m$ perpendicular to a magnetic field of $0.625 \ T$. Find the de Broglie wavelength associated with the proton in $\mathring{A}$.

If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ $(E_0 = \text{constant} > 0)$, with an initial velocity $\vec{V} = v_0\hat{i}$ $(v_0 = \text{constant} > 0)$. If $\lambda_0 = \frac{h}{mv_0}$, its de Broglie wavelength at time $t$ is . . . . . . .