Which of the following particles has the shortest de-Broglie wavelength, assuming they have the same kinetic energy?

The ratio of the de-Broglie wavelength of an $\alpha$-particle to that of a proton,when both are subjected to the same magnetic field such that the radii of their paths are equal,assuming the magnetic field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton,is:

Crystal diffraction experiments can be performed using $X-$rays,or electrons accelerated through an appropriate voltage. Which probe has greater energy? (For quantitative comparison,take the wavelength of the probe equal to $1\,\mathring{A}$,which is of the order of inter-atomic spacing in the lattice) $(m_{e}=9.11 \times 10^{-31}\,kg)$

An $\alpha$-particle moves in a circular path of radius $0.83\, cm$ in the presence of a magnetic field of $0.25\, Wb/m^2$. The de Broglie wavelength associated with the particle will be .............. $\mathring{A}$.

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is the mass of the electron and $m_p$ is the mass of the proton).

If the kinetic energy of an electron and a proton are equal,find the ratio of their associated de Broglie wavelengths.