JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Mosley's Law

Mosley studied the characteristic X-ray spectrum of a number of a heavy elements and concluded that the spectra of different elements are very similar and with increasing atomic number, the spectral lines merely shift towards higher frequencies.

He also gave the following relation \[\sqrt{\nu }=a\,(Z-b)\]

where v = Frequency of emitted line, Z = Atomic number of target, a = Proportionality constant, b = Screening constant or Shielding constant.

(Z - b) = Effective atomic number

a and b doesn't depend on the nature of target. Different values of b are as follows

b = 1                           for          K-series

b = 7.4                       for          L-series

b = 19.2                     for          M-series

(1) Mosley's law supported Bohr's theory

(2) It experimentally determined the atomic number (Z) of elements.

(3) This law established the importance of ordering of elements in periodic table by atomic number and not by atomic weight.

(4) Gaps in Moseley's data for A = 43, 61, 72, 75 suggested existence of new elements which were later discovered.

(5) The atomic numbers of Cu, Ag and Pt were established to be 29, 47 and 78 respectively.

(6) When a vacancy occurs in the K-shell, there is still one electron remaining in the K-shell. An electron in the L-shell will feel an effective charge of (Z - 1)e due to + Ze from the nucleus and ? e from the remaining K-shell electron, because L-shell orbit is well outside the K-shell orbit.

(7) Wave length of characteristic spectrum \[\frac{1}{\lambda }=R{{(Z-b)}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] and energy of X-ray radiations. \[\Delta E=h\nu =\frac{hc}{\lambda }=Rhc{{(Z-b)}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\]

(8) If transition takes place from \[{{n}_{2}}=2\] to \[{{n}_{2}}=1\] (\[{{K}_{\alpha }}\]- line)

(i) \[a=\sqrt{\frac{3RC}{4}}=2.47\times {{10}^{15}}Hz\]

(ii) \[{{\nu }_{K\alpha }}=RC{{(Z-1)}^{2}}\left( 1-\frac{1}{{{2}^{2}}} \right)=\frac{3RC}{4}{{(Z-1)}^{2}}\] \[=2.47\times {{10}^{15}}{{(Z-1)}^{2}}Hz\]

(iii) In general the wavelength of all the K-lines are given by \[\frac{1}{{{\lambda }_{K}}}=R{{(Z-1)}^{2}}\,\left( 1-\frac{1}{{{n}^{2}}} \right)\] where n = 2, 3, 4, ?.

While for \[{{K}_{\alpha }}\]  line \[{{\lambda }_{K\alpha }}=\frac{1216}{{{(Z-1)}^{2}}}{\AA}\]

(iv) \[{{E}_{K\alpha }}=10.2{{(Z-1)}^{2}}eV\]