When the kinetic energy of an electron is increased,the wavelength of the associated wave will

The de Broglie wavelength of a neutron at $27^{\circ}C$ is $\lambda$. What will be its de Broglie wavelength at $927^{\circ}C$?

Find the de Broglie wavelength of an electron in a metal at $27^{\circ}C$.

An $\alpha$-particle moves in a circular path of radius $1 \ cm$ in a uniform magnetic field of $0.125 \ T$. The de Broglie wavelength associated with the $\alpha$-particle is

What does the de Broglie equation for an electron represent?

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: