The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of the electron becomes $\frac{E}{4}$,then its de Broglie wavelength will be:

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

$A$ photon and an electron are given the same energy $(10^{-20} \ J)$. If the wavelengths associated with the photon and the electron are $\lambda_{Ph}$ and $\lambda_{el}$ respectively,then which of the following statements is correct?

If the kinetic energy of a particle is increased to $16$ times, the percentage change in the de Broglie wavelength of a particle is (in $\%$)

$(a)$ For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40 \times 10^{-10} \; m ?$
$(b)$ Also find the de Broglie wavelength of a neutron,in thermal equilibrium with matter,having an average kinetic energy of $\frac{3}{2} k T$ at $300 \; K$.

If the de Broglie wavelength of an electron is $5200 \ \mathring{A}$,what is its velocity?