If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is

$A$ proton is accelerated through a potential difference of $100 \ V$. To have the same de Broglie wavelength,what potential difference must be applied across a doubly ionized $_4^8 Be$ nucleus (in $V$)?

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 kinetic energy of an electron and a proton is $10^{-32} \ J$. Then the relation between their de-Broglie wavelengths is

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

The figure shows four situations in which an electron is moving in an electric or magnetic field. In which case is the de Broglie wavelength of the electron increasing?