What will be the ratio of de-Broglie wavelengths of a proton and an $\alpha$-particle of the same energy?

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12} \ m$, the minimum electron energy required is close to .............. $keV$.

After absorbing a slowly moving neutron of mass $m_{N}$ (momentum $0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_{1}$ and $5m_{1}$ $(6m_{1} = M + m_{N})$ respectively. If the de-Broglie wavelength of the nucleus with mass $m_{1}$ is $\lambda$,then the de-Broglie wavelength of the other nucleus will be:

Of the following particles having the same kinetic energy,the one which has the largest de Broglie wavelength is

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$,their de Broglie wavelength in the frame of reference attached to their centre of mass is

An electron accelerated by a potential difference $V$ has a de-Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $9V$,its de-Broglie wavelength will be