What is the ratio of the de Broglie wavelengths of a deuteron and a proton accelerated through the same potential difference?

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be:

The interatomic spacing in a crystal is $1.227 \ \mathring A$. What is the maximum order of diffraction for electrons accelerated by $10 \ kV$?

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 .............. $\mathring{A}$.

An electron and a proton are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{p}$ to $\lambda_{e}$ is $[m_{e} = \text{mass of electron}, m_{p} = \text{mass of proton}]$

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is