Radius of $${n^{th}}$$  orbit is given by $${r_n} = \frac{{{r_0} \times {n^2}}}{Z}$$
For  $$_3L{i^{2 + }},r = \frac{{{r_0}}}{3} = \frac{{0.53}}{3} = 0.176\mathop {\text{A}}\limits^{\text{o}} $$

The electronic configuration of $$Cu (29)$$  is an exceptional case due to exchange of energy and symmetrical distribution of electrons in orbital to acquire more stability.

$$ = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^1}$$

The number of orbitals in a subshell $$ = \left( {2l + 1} \right)$$
where, $$l$$ = azimuthal quantum number
Since, each orbital contains maximum two electrons, the number of electrons in any subshell
$$\eqalign{ & = 2 \times {\text{number of orbitals}} \cr & = 2\left( {2l + 1} \right) \cr & = 4l + 2 \cr} $$

Let relative abundance of $$Cl{\text{ - 37}} = x\% $$
then relative abundance of $$Cl{\text{ - 35}} = \left( {100 - x} \right)\% $$
Average atomic mass $$ = \frac{{x \times 37 + \left( {100 - x} \right)35}}{{100}} = 35.5$$
$$\eqalign{ & \Rightarrow 37x + 3500 - 35x = 3550 \Rightarrow x = 25 \cr & \therefore 100 - x = 75 \cr} $$
Thus, the ratio of $$Cl$$ - 37 and $$Cl$$ - 35 is $$x:\left( {100 - x} \right){\text{ = 25 : 75 = 1 : 3}}$$

In the photoelectric effect, the energy of the emitted electron is smaller than that of the incident photon because some energy of photon is used to eject the electron and remaining energy is used to increase the kinetic energy of ejected electron.

The smallest value of energy of an electron in $$H$$ atom in ground state can absorb is $$ = {E_2} - {E_1}.$$
$$\eqalign{ & = \frac{{ - 13.58}}{4} - \left( {\frac{{ - 13.58}}{1}} \right) \cr & = 10.19 \cr} $$

In Balmer series, the lines appear in visible region.

The wavelength of light is related to its energy by the equation, $$E = \frac{{hc}}{\lambda },\left( {E = hv} \right)$$
$${\text{Given,}}\,\,\lambda = 45nm = 45 \times {10^{ - 9}}m$$      $$\left[ {\because \,\,1nm = {{10}^{ - 9}}m} \right]$$
$${\text{Hence,}}\,\,E$$   $$ = \frac{{6.63 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{45 \times {{10}^{ - 9}}m}}$$
                 $$ = 4.42 \times {10^{ - 18}}J$$
Hence, the energy corresponds to light of wavelength $$45\,nm$$  is $$4.42 \times {10^{ - 18}}J.$$

Total number of nodes $$ = n - 1$$
For $$d$$ - orbital, radial nodes $$ = n - l - 1 = n - 3$$     and there are 2 angular nodes.

No. of neutrons = Mass number - Atomic number
                         = 70 - 30 = 40.