What is the de Broglie wavelength of an electron accelerated through a potential difference of $ 100 \ V $ (in $\text{Å}$)?

An electron is accelerated through a potential difference of $10,000 \ V$. Its de-Broglie wavelength is nearly: $(m_{e} = 9 \times 10^{-31} \ kg)$

The kinetic energy of an electron having de-Broglie wavelength $\lambda$ is ($h=$ Planck's constant,$m=$ mass of electron).

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 . . . . . . .

The proton and $\alpha$-particle are accelerated through the same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$-particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton).

If a photon,an electron,and a uranium nucleus have the same de Broglie wavelength,which one will have the highest energy?