Atoms

$=5\mathrm{M}\left({\mathrm{u}}^2-\frac{119{\mathrm{G}\mathrm{M}}_{\mathrm{e}}}{200\mathrm{R}}\right)$

$T=\frac{2\pi }{\omega }=2\pi \frac{r}{v}\propto \frac{n^2}{\frac{1}{n}}\propto n^3$

$\therefore \mathrm{}v=\frac{1}{\mathrm{T}}\propto \frac{1}{{\mathrm{n}}^3}$

$\frac{v_2}{v_1}=\frac{1^3}{2^3}=\frac{1}{8}$

E_K - E_L = hc/λ_Kα
E_K - E_L = (4.14*10? ¹? * 3*10? )/0.071*10? = 17500 eV = 17.5 keV.
E_L = E_K - 17.5 = 27.5 - 17.5 = 10 keV.

Energy of electron = 3 eV
It forms H atom in n=3 state. Energy released E = 3 - (-13.6/9) = 3 + 1.51 = 4.51 eV.
Photon Energy = 4.51 eV
Threshold energy = hc/λ = 12400eVÅ / 4000Å = 3.1 eV.
kE_max = 4.51 - 3.1 = 1.41 eV

f = (N? - N? (e? )/N?
f = 1 - e?
df/dt = 0 - e? * -λ = λe?

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=13.6*\frac{24}{25}eV$

$\Rightarrow \frac{hc}{\lambda }=13.6*\frac{24}{25}eV..........\left(1\right)$

With the help of conservation of linear momentum, we can write

$\frac{h}{\lambda }=m_H{v}_H\Rightarrow \frac{hc}{\lambda }=c{m}_H{v}_H\Rightarrow v_H=\frac{\frac{hc}{\lambda }}{c{m}_H}=\frac{13.6*\frac{24}{25}*1.6*10^{-19}}{3*10^8*1.67*10^{-27}}=4.17m/s$

V = 1.24 * 10⁶ volt

${\lambda }_{min}=?$

${\lambda }_{min}=\frac{hc}{eV}=\frac{1242nm}{e\left(1.24*10^6\right)}=\frac{1000*10^{-9}}{10^6}\Rightarrow {\lambda }_{min}=10^{-3}nm$

Energy difference $\Delta E=\frac{hc}{\lambda }$

$\therefore \lambda \propto \frac{1}{\Delta E}$

${(\Delta E)}_{6-2}>{(\Delta E)}_{5-2}>{(\Delta E)}_{4-2}>{(\Delta E)}_{3-2}$

${\lambda }_{6-2}<{\lambda }_{5-2}<{\lambda }_{4-2}<{\lambda }_{3-2}$
 

At same temperature, curve with higher volume corresponds to lower pressure.

$V_3>V_2>V_1$

$\Rightarrow P_1>P_2>P_3$

(We draw a straight line parallel to volume axis to get this)

Statement I is true as atoms are electrically neutral because they contain equal number of positive and negative charges.

Statement II is wrong as atom of most of the elements are stable and emit characteristic spectrum. But this statement is not true for every atom.