$A$ proton is accelerated through a potential difference of $100 \ V$. To have the same de Broglie wavelength,what potential difference must be applied across a doubly ionized $_4^8 Be$ nucleus (in $V$)?

An electron of mass $m$ is accelerated through a potential difference of $V$. Its de Broglie wavelength is $\lambda$. If a proton of mass $M$ is accelerated through the same potential difference,its de Broglie wavelength will be ..............

The de-Broglie wavelength of an electron having $80 \ eV$ of energy is nearly .............. $\mathring{A}$ ($1 \ eV = 1.6 \times 10^{-19} \ J$,Mass of electron $= 9 \times 10^{-31} \ kg$,Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$).

$A$ particle of mass $2 \times 10^{-27} \,kg$ has a de-Broglie wavelength of $3.3 \times 10^{-10} \,m$. The kinetic energy of this particle is (Planck's constant $h = 6.6 \times 10^{-34} \,J \cdot s$).

What is the de Broglie wavelength of an electron accelerated through $80 \ V$ in $\mathring A$?

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).