For a nucleus $${ }_{\mathrm{A}}^{\mathrm{A}} \mathrm{X}$$ having mass number $$\mathrm{A}$$ and atomic number $$\mathrm{Z}$$

A. The surface energy per nucleon $$\left(b_{\mathrm{s}}\right)=-a_{1} A^{2 / 3}$$.

B. The Coulomb contribution to the binding energy $$\mathrm{b}_{\mathrm{c}}=-a_{2} \frac{Z(Z-1)}{A^{4 / 3}}$$

C. The volume energy $$\mathrm{b}_{\mathrm{v}}=a_{3} A$$

D. Decrease in the binding energy is proportional to surface area.

E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. ( $$a_{1}, a_{2}$$ and $$a_{3}$$ are constants)

Choose the most appropriate answer from the options given below:

A small particle of mass $$m$$ moves in such a way that its potential energy $$U=\frac{1}{2} m ~\omega^{2} r^{2}$$ where $$\omega$$ is constant and $$r$$ is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of $$n^{\text {th }}$$ orbit will be proportional to,

The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is :

An electron of a hydrogen like atom, having $$Z=4$$, jumps from $$4^{\text {th }}$$ energy state to $$2^{\text {nd }}$$ energy state. The energy released in this process, will be :

(Given Rch = $$13.6~\mathrm{eV}$$)

Where R = Rydberg constant

c = Speed of light in vacuum

h = Planck's constant