If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

An electron accelerated by a potential difference $V$ has a de-Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $9V$,its de-Broglie wavelength will be

The resolving power of an electron microscope is greater than the resolving power of an optical microscope because

The potential difference $V$ required for accelerating an electron to have the de-Broglie wavelength of $1 \text{ Å}$ is (in $\text{ V}$)

In an electron microscope,which nature of the electron is used?

The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy $K$ of the electron and the photon,which of the following presents the correct relation between the de Broglie wavelengths of the two?