Structure of Atom

According to Heisenberg uncertainty principle

$\Delta p.\Delta x \ge {h \over {4\pi }}$ or m$\Delta v.\Delta x \ge {h \over {4\pi }}$

$ \Rightarrow $ (m$\Delta v$)² $\ge {h \over {4\pi }}$ ($ \because $ $\Delta x$ = $\Delta p$)

$ \Rightarrow $ $\Delta v \ge$ ${1 \over {2m}}\sqrt {{h \over \pi }} $

(i) represents an electron in 3s orbital

(ii) is not possible as value of m varies from 0, 1, .... ($n$ -1)

(iii) represents an electron in 4f orbital

(iv) is not possible as value of m varies from -$l$ ... +$l$

(v) is not possible as value of m varies from -$l$ ... +$l$, it can never be grater than $l$

Magnetic quantum number describes the orientation.

$\Delta x.m\Delta v = h/4\pi $

0.1 $ \times $ 10^-10 $ \times $ 9.11 $ \times $ 10^-31 $ \times $ $\Delta v$ = ${{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143}}$

$ \therefore $ $\Delta v$ = ${{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143 \times 0.1 \times {{10}^{ - 10}} \times 9.11 \times {{10}^{ - 31}}}}$

= 5.79 $ \times $ 10⁶ m s^-1

E_n = -K${\left( {{Z \over n}} \right)^2}$

Z = 1; n = 2

E₂ = ${ {{-K \times 1 \over 4}}}$$ \Rightarrow $ E₂ = -328 kJ mol^-1;

K = 4 $ \times $ 328

E₄ = ${ {{-K \times 1 \over 16}}}$ $ \Rightarrow $ E₄ = -4 $ \times $ 328 ${ {{ 1 \over 16}}}$

= -82 kJ mol^-1

E = h$v$ or $v$ = E/h

For H atom, E = ${{ - 21.76 \times {{10}^{ - 19}}} \over {{n^2}}}J\,at{m^{ - 1}}$

$\Delta E = - 21.76 \times {10^{ - 19}}\left( {{1 \over {{4^2}}} - {1 \over {{1^2}}}} \right)$

= 20.40 $ \times $ 10^-19 J atm^-1

$v$ = ${{20.40 \times {{10}^{ - 19}}} \over {6.626 \times {{10}^{ - 34}}}}$ = 3.079 $ \times $ 10¹⁵ s^-1

$v$ = c/$\lambda $

$\lambda $ = ${c \over v}$ = ${{3 \times {{10}^8}} \over {8 \times {{10}^{15}}}}$ = 37.5 $ \times $ 10^-9 m

= 37.5 nm $ \approx $ $4 \times {10^1}$ nm

K.E = 1/2 mv²

= ${\left( {{{\pi {e^2}} \over {nh}}} \right)^2} \times 2m$ [ $ \because $ v = ${{2\pi {e^2}} \over {nh}}$]

$ \therefore $ Total energy = E_n = ${{ - 2{\pi ^2}m{e^4}} \over {{n^2}{h^2}}}$

= -${\left( {{{\pi {e^2}} \over {nh}}} \right)^2}$ $ \times $ 2m = -K.E

$ \therefore $ K.E = - E_n

Energy of first excited state is -3.4 eV

$ \therefore $ Kinetic energy of the same orbit (n = 2) will be +3.4 ev

n = 3, l = 2, m = +2

It shows one of the five d-orbitals(3d)

For $\pi $ overlap, the lobes of the atomic orbitals are perpendicularto the line joining the nuclei.

Hence only sidewise overlapping takes place.

No. of electrons CO = CN^- = NO⁺ = $C_2^{2 - }$ = 14, So these are isoelectronics.

E = ${{hc} \over \lambda }$

$ \Rightarrow \lambda $ = ${{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {3.03 \times {{10}^{ - 19}}}}$

= 656 nm