Structure of Atom

Before 20th century, atoms were widely considered the indivisible building blocks of matter. This view goes back to Indian and Greek philosophies, as old as 400 B.C. It also became one established thought on a scientific basis by John Dalton in 1808. With his theory, several fundamental laws of chemistry were established. But those laws failed against observations like static electricity. The experimental observations made towards the end of the 19th and beginning of the 20th century definitively proved that atoms are made of subatomic particles. 

$E_n=-R_H\left(\frac{Z^2}{n^2}\right)J$

For ${}^{+}\left(n=1\right)$ ,

$E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4R_H$

$\therefore R_H=\frac{x}{4}$

For ${}^{3+}\left(n=2\right)$ ,

$E_n=-R_H\left(\frac{z^2}{n^2}\right)J$

$=-\frac{x}{4}*\left(\frac{4*4}{2*2}\right)=-xJ$

Kindly go through the solution

Value of n &  $\mathcal{l}$ can't be same

Energy of 1 photon = $\frac{hc}{\lambda }$  

Energy of 1 mole of photon =  $N_A*\frac{hc}{\lambda }$  

$=\frac{6.022*10^{23}*6.63*10^{-34}*3*10^8}{300*10^{-9}}$  

= 0.399 * 10⁶ J = 399 KJ

Only tritium is the isotope of hydrogen which is radioactive in nature that emits low energy b^- particles since its $\frac{n}{p}$ ratio is above stability belt.

Electronic configuration of Ga⁺ ion = [Ar] 3d¹⁰ 4s² 4p⁰

Last electron goes into S-orbital, hence

Azimuthal quantum number $\left(\mathcal{l}\right)$ for last electron = 0

Zn (30) = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$  

$Z{n}^{+}=\left[Ar\right]4s^{1}3d^{10}$      

Outermost electron is 4s electron,

$n=4,\mathcal{l}=0,m=0,s=\pm 1/2$         

Ge (Z = 32)

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

$m_{\mathcal{l}}=0$       (for 4s, 3s, 2s, 1s) 4 orbital

$m_{\mathcal{l}}$       = 0 (one p orbital of 2p and 3p)

$m_{\mathcal{l}}$ = 0 (one d orbital)

Total orbitals = 7

Ans. = 7

mvr = $\frac{nh}{2\pi }\left(angularmomentum\right)$  

$KE=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}$ (Bohr's kinetic energy)

$=\frac{4h^2}{8{\pi }^{2}m{\left(4a_0\right)}^2}\left(n=2\right)$

$K.E.=\frac{1}{32{\pi }^2}\frac{h^2}{m{a}_0^2}$

Comparing with $\frac{h^2}{xm{a}_0^2}$

x = 32π² = 315.50

10x = 3155

According to Bohr's model, velocity of electron increases according to atomic number (z).

$v_e\propto \frac{z}{n}$ (z = Atomic number and n = number of shell/principal quantum number)

Similarly, velocity of electron decreases according to number of shell (i.e. principal quantum number).