Isoelectronic species means species having same number of electrons.
In $$CO;$$  no. of electrons = 6 + 8 = 14
In $${N_2};$$  no. of electrons =7 +7= 14

$$F{e^{ + + }}\left( {26 - 2 = 24} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^6}$$         hence no. of d electrons retained is 6.
[Two $$4s$$ electron are removed]

$$\lambda = \frac{h}{{mv}} = \frac{{6.6 \times {{10}^{ - 34}}}}{{60 \times {{10}^{ - 3}} \times 10}} = {10^{ - 33}}m$$

$$\lambda = \frac{h}{{mv}}$$   if $$\nu $$  is same, then higher the value of $$m$$  means lower will be the value of $$\lambda .$$

TIPS/Formulae :
$$\eqalign{ & {r_n} = 0.529\frac{{{n^2}}}{Z}\,\mathop {\text{A}}\limits^{\text{o}} \cr & {\text{For hydrogen,}}\,\,n = 1\,\,and\,Z = 1; \cr & \therefore \,\,\,{r_H} = 0.529 \cr & {\text{For}}\,B{e^{3 + }},n = 2\,\,{\text{and}}\,\,Z = 4\,; \cr & \therefore \,\,{r_{B{e^{3 + }}}} = \frac{{0.529 \times {2^2}}}{4} = 0.529 \cr} $$

$$n = 3,$$  $$l = 1$$  represents $$3p$$  orbital. Since $$p$$  has three orientations $${p_x},{p_y}$$  and $${p_z},6$$  electrons will show same quantum number values of $$n$$  and $$l.$$

$$\eqalign{ & \frac{1}{\lambda } = R{Z^2}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] \cr & = R \times {2^2}\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right] \cr & \Rightarrow 3R;\,\lambda = \frac{1}{{3R}} \cr} $$

$$3{d^5}4{s^1}$$ system is more stable than $$3{d^4}4{s^2},$$   hence former is the ground state configuration.

The main postulates of Bohr model of atom are
(i) The electrons in an atom revolve around the nucleus only in certain selected circular paths, called orbits.
(ii) The energy is emitted or absorbed only when the electrons jump from one energy level to another.
(iii) Only those orbits are permitted in which the angular momentum of the electron is a whole number multiple of $$\frac{h}{{2\pi }}$$  ( where, $$h$$ is Planck's constant ) that's why only certain fixed orbits are allowed, i.e. the momentum of an electron is quantised.

$$\eqalign{ & {\text{Maximum number of emission lines}} \cr & \frac{{n\left( {n - 1} \right)}}{2} \cr & = \frac{{6\left( {6 - 1} \right)}}{2} \cr & = \frac{{30}}{2} \cr & = 15\,\,{\text{lines}} \cr} $$