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)

For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : The density of the copper $(^ {64}_{29} \text{Cu})$ nucleus is greater than that of the carbon $(^ {12}_{6} \text{C})$ nucleus.

Reason (R) : The nucleus of mass number A has a radius proportional to $ A^{1/3} $.

In the light of the above statements, choose the most appropriate answer from the options given below :