$A$ particle has mass $m$. If its kinetic energy is increased to four times its initial value,what will be the change in its de Broglie wavelength?

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

An electron (mass $m$) with an initial velocity $\overrightarrow{v} = v_{0} \hat{i} \left(v_{0} > 0\right)$ is moving in an electric field $\overrightarrow{E} = -E_{0} \hat{i} \left(E_{0} > 0\right)$ where $E_{0}$ is constant. If at $t = 0$ the de Broglie wavelength is $\lambda_{0} = \frac{h}{mv_{0}}$,then its de Broglie wavelength after time $t$ is given by:

The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$. What will be the wavelength of the particle if its kinetic energy becomes $\frac{K}{4}$?

If a proton is accelerated through a potential difference of $1000 \,V$, then its de-Broglie wavelength is (given, $m_p = 1.67 \times 10^{-27} \,kg$, $h = 6.63 \times 10^{-34} \,J-s$).

Find the de Broglie wavelength associated with a Helium $(He)$ atom in Helium gas at room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure.