$A$ proton and an $\alpha$-particle are accelerated from rest by $2\,V$ and $4\,V$ potentials, respectively. The ratio of their de-Broglie wavelength is: (in $:1$)

$A$ particle of mass $1 \, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6 \, m/s$. The velocity of the particle is (mass of electron $= 9.1 \times 10^{-31} \, kg$).

$A$ ball of mass $0.12 \ kg$ is moving with a speed $20 \ m \ s^{-1}$. Then its de Broglie wavelength is . . . . . . . ( $h = 6.63 \times 10^{-34} \ J \ s$ )

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

$A$ stationary mass $M$ splits into two fragments of masses $m_1$ and $m_2$. What is the ratio of their de Broglie wavelengths,${\lambda _1}/{\lambda _2}$?

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....... $KeV$.