The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of the electron becomes $\frac{E}{4}$,then its de Broglie wavelength will be:

The wavelength associated with an electron accelerated through a potential difference of $100 \ V$ is nearly .............. $ \mathring A $

The de Broglie wavelength associated with a proton under the influence of an electric potential of $100 \ V$ is

The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of the proton,$\lambda_2$ is the wavelength associated with the photon,and the energy of the photon is $E$,then $(\lambda_1 / \lambda_2)$ is proportional to

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 $:-$

If an electron has an energy such that its de-Broglie wavelength is $5500 \ \text{Å}$,then the energy value of that electron is $(h = 6.6 \times 10^{-34} \ \text{Js}, m_e = 9.1 \times 10^{-31} \ \text{kg})$.