Physics Atoms

Using Bohr's theory :

$\lambda =\frac{hc}{E_2-E_1}=\frac{1242eV-nm}{\left(-3.4\right)-\left(-13.6\right)}=121.8nm$

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)

For first line of Lyman

$\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=R\frac{3}{4}$

$\lambda =\frac{4}{3R}$

3rd line (Paschen)

$\frac{1}{{\lambda }_3}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right)=\frac{R}{9}*\frac{3}{4}$

2nd line (B almer)

${\frac{1}{\lambda }}_2=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{R}{4}*\frac{3}{4}$

$a\left(\frac{4}{3R}\right)=\frac{20}{3R}\Rightarrow a=5$

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.

According to de-Broglie's hypothesis

$\lambda =\frac{h}{mv}\Rightarrow p=mv=\frac{h}{\lambda }\Rightarrow K=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$

Cut-off wavelength of emitted X-ray = $\frac{hc}{K}=hc*\frac{2m{\lambda }^2}{h^2}=\frac{2mc{\lambda }^2}{h}$

Energy of electron in first existed state will be -3.4 eV.

So total energy difference will be (2.6 + 3.4) eV.

Wavelength $\left(\lambda \right)=\frac{1242eV-nm}{6eV}=207nm$

$Frequency=\frac{c}{\lambda }=\frac{3*10^8}{207*10^{-9}}=1.45*10^{9}MHz$

No. of different wavelengths

$=\frac{n*\left(n-1\right)}{2}=\frac{6*5}{2}=15$

$h{f}_1=R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R*\frac{8}{9}$

$h{f}_2=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R*\frac{3}{4}$

$\frac{f_1}{f_2}=\frac{8/9}{3/4}=\frac{32}{27}$

f₂ = f₁ x $\left(\frac{27}{32}\right)$

$=\left(2.92*10^{15}\right)*\left(\frac{27}{32}\right)$

$=2.46*10^{15}Hz$

  $v\propto \frac{z}{n}$ $\frac{}{}$

$\frac{}{}$  $v=k\frac{z}{n}$ [K constant]

for 3rd orbit of He+

${}_{}$ $v_{H{e}^{+}}=\frac{k*2}{3}=\frac{2k}{3}$  - (1)

$$ $v_H=\frac{k*1}{3}=\frac{k}{3}$  - (2)

$\frac{v_{H{e}^{+}}}{v_H}=2:1$