If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is:

$A$ particle of mass $8 \mu g$ in motion collides with another stationary particle of mass $4 \mu g$. If the collision is perfectly elastic and one-dimensional, the ratio of their de Broglie wavelengths after collision is (in $2 : 1$)

If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is

The de Broglie wavelength of an electron moving with a velocity $1.5 \times 10^8 \, m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is

What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?