An electron accelerated through a potential of $10000 \ V$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential, so that the wavelength is doubled (in $V$)?

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:

An $\alpha$-particle and a deuteron are moving with velocities $v$ and $2v$ respectively. What will be the ratio of their de Broglie wavelengths?

If the de Broglie wavelength of an electron is $5200 \ \mathring{A}$,what is its velocity?

An $\alpha$-particle and a proton are accelerated from rest by the same potential,then the ratio of their de-Broglie wavelength is

What is the de Broglie wavelength associated with
$(a)$ an electron moving with a speed of $5.4 \times 10^{6} \; m/s$,and
$(b)$ a ball of mass $150 \; g$ travelling at $30.0 \; m/s$?