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

$A$ photon and an electron have equal energy $E$. The ratio of $\lambda(\text{electron})$ to $\lambda(\text{photon})$ is proportional to

The de Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths is . . . . . . .

$A$ proton and an $\alpha-$particle are accelerated through a potential difference of $100 \ V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha-$particle is:

When the kinetic energy of an electron is increased,the wavelength of the associated wave will

Two large parallel plates are connected to a $100 \, V$ power supply. These plates have a fine hole at the centre. An electron having energy $200 \, eV$ is directed such that it passes through the holes. When it emerges, its de-Broglie wavelength is ............. $\mathring{A}$