Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will

$A$ particle of charge $q$,mass $m$,and energy $E$ has de-Broglie wavelength $\lambda$. For a particle of charge $2q$,mass $2m$,and energy $2E$,the de-Broglie wavelength is:

The idea of matter waves was given by

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through the same potential of $100\, V$. What should nearly be the ratio of their wavelengths? $(m_{P} = 1.00727\, u, m_{e} = 0.00055\, u)$

$A$ particle is moving $5$ times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 \times 10^{-4}$. The mass of the particle is close to

$A$ particle of charge $q$ and mass $m$ enters a region of a transverse electric field of $E_{0} \hat{j}$ with initial velocity $v_{0} \hat{i}$. The time taken for the change in the de-Broglie wavelength of the charge from the initial value of $\lambda_{0}$ to $\lambda_{0} / 3$ is proportional to