The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)

$A$ proton and an alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of the proton to that of the alpha particle will be (mass of alpha particle is four times the mass of the proton,and the charge of the alpha particle is twice the charge of the proton).

An electron of mass $m$ with an initial velocity $\overrightarrow{v}=v_0 \hat{i}$ $(v_0>0)$ enters an electric field $\overrightarrow{E}=-E_0 \hat{k}$. If the initial de Broglie wavelength is $\lambda_0$,the value after time $t$ would be $:-$

$A$ nucleus of mass $M$,moving slowly,absorbs a neutron of mass $m_N$ and then breaks into two nuclei of masses $m_1$ and $5m_1$. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then what will be the de Broglie wavelength of the other nucleus?

The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$. What will be the wavelength of the particle if its kinetic energy becomes $\frac{K}{4}$?

$A$ 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