Structure of Atom

(A) Cr = [Ar]3d⁵4s¹

              (B) m =   $-\mathcal{l}to+\mathcal{l}$

              (C) According to Aufbau principle, orbital are filled in order of their increasing energies.

              (D) Total nodes = n – 1

 Number of neutrons = 26

Number of electrons = 25

% of extra electrons = $\frac{26-25}{25}*100=4\mathrm{\%}$

 $\lambda =\frac{h}{\sqrt[]{2mE}}E=KineticEnergy$

$E=\frac{h^2}{2m{\lambda }^2}$

$=\frac{{\left(6.626*10^{-34}\right)}^2}{2*9.1*10^{-31}*{\left(3.3*10^{-10}\right)}^2}$

= 2.215 * 10^-18

Eabsorbed – Erequired + K. E

$\frac{Eabsorbed}{Erequired}=1+\frac{K.E}{Erequired}$

= $=1+\frac{2.215*10^{-18}}{13.6*1.602*10^{-19}}=2.016$

$\approx$ 2

n = 1, 2, 3 …….;   $\mathcal{l}=0$ ……. to n – 1, $m_{\mathcal{l}}=\mathcal{l}.......0......+\mathcal{l}$  

A. n = 3   $\mathcal{l}=3$ $m_{\mathcal{l}}=-3$  is incorrect as $\mathcal{l}$  can not be equal to n.

B. n = 3   $\mathcal{l}=2$   $m_{\mathcal{l}}$ = -2 is correct set.

C. n = 2   $\mathcal{l}=1$ $m_{\mathcal{l}}$ = +2 is incorrect set as  $\mathcal{l}$ can not be equal to n.

So; correct set of quantum numbers is B and C.

According to the de Broglie equation, the particles like electrons have wave-like properties. The quantum mechanical model foundation is laid by this concept and it also explains the stability of electron orbits using wave behavior.

The hydrogen atom can be accurately explained by Bohr's model. However, it does not account for shielding effects, electron-electron interactions, and the wave nature of electrons for multi-electron atoms.

Quantum numbers describe the unique position and energy of an electron in an atom. These are a set of four numbers and are important for predicting chemical behavior and understanding the distribution of electrons in orbitals.

λ₁ = 589 nm = 589 x 10^-9 m

ν₁ = c / λ₁ = (3 x 10⁸ ms^-1) / (589 x 10^-9 m) = 5.0934 x 10¹⁴ s^-1

λ₂ = 589.6 nm = 589.6 x 10^-9 m

ν₂ = c / λ₂ = (3 x 10⁸ ms^-1) / (589.6 x 10^-9 m) = 5.0882 x 10¹⁴ s^-1

ΔE = E₁ – E₂ = h [ν₁ – ν₂]

= (6.626 x 10^-34Js) x [ (5.0934 x 10¹⁴ s^-1) – (5.0882 x 10¹⁴ s^-1)

= 3.31 x 10^-22 J

(i) Energy of photon (E) = hc / λ

= (6.626 * 10^-34 Js) x (3 * 10⁸ m s^-1) / (4 * 10^-7 m) = 4.969 x 10^-19 J

Since, 1.6020 * 10^-19 J= 1 eV

So, 1 J= (1 eV) / (1.6020 * 10^-19 J)

Hence, 4.969 x 10^-19 J = (1eV) x (4.969 x 10^-19 J) / (1.602 x 10^-19 J) = 3.1 eV

(ii) Kinetic energy of emission = Energy – work function

= (3.1 – 2.13) = 0.97 eV

(iii) Kinetic energy of emission = 0.97 eV

=> ½ mv²= 0.97 eV = 0.97 x 1.602 x 10^-19 J = 0.97 x 1.602 x 10^-19 kg m² s^-2

=> v² = (2 x 0.97 x 1.602 x 10^-19 kg m² s^-2 ) / (9.1 x 10^-31 kg) = 0.34 x 10¹² m² s^-2

=> v = (0.34 x 10¹² m² s^-2)^1/2 = 0.583 x 10⁶ ms^-1 = 5.83 x 10⁵ ms^-1