Which of the following particles has the shortest de-Broglie wavelength, assuming they have the same kinetic energy?

$A$ photon and an electron are given the same energy $(10^{-20} \ J)$. If the wavelengths associated with the photon and the electron are $\lambda_{Ph}$ and $\lambda_{el}$ respectively,then which of the following statements is correct?

We wish to see inside an atom. Assuming the atom to have a diameter of $100 \ pm$, this means that one must be able to resolve a width of say $10 \ pm$. If an electron microscope is used, the minimum electron energy required is about....... $KeV$.

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is the mass of the electron and $m_p$ is the mass of the proton).

An electron beam used in an electron microscope,when accelerated by a voltage of $20\,kV$,has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to $40\,kV$,then the de-Broglie wavelength associated with the electron beam would be:

At room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure,the de Broglie wavelength associated with a helium atom is compared with the average distance between two atoms. Which of the following relations is correct?