The minimum frequency of photon required to break a particle of mass 15.348 amu into $4 \alpha$ particles is $\_\_\_\_$ kHz .

[mass of He nucleus = $4.002 \mathrm{amu}, 1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ ]

$7.9 \mathrm{MeV} \alpha$-particle scatters from a target material of atomic number 79 . From the given data the estimated diameter of nuclei of the target material is (approximately) $\_\_\_\_$ m.

$$ \left[\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2 \text { and electron charge }=1.6 \times 10^{-19} \mathrm{C}\right] $$

The energy of an electron in an orbit of the Bohr's atom is $-0.04E_0$ eV where $E_0$ is the ground state energy. If $L$ is the angular momentum of the electron in this orbit and $h$ is the Planck's constant, then

$ \frac{2\pi L}{h} $ is ________ :

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is $\_\_\_\_$ m. (Atomic number of gold $=79$ and $\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^9$ in SI units)