A,B and C are iso-diaphers while C,D and E are isobars. Calculate the difference of protons between A and ${\mathrm{E}}_{82}^{206}$ A→B→C→D→E. Given : Iso-diaphers and isobars are formed in Successive the α and ß number of and respectively x and y. Then x+y value is 

• Isodiaphers have the same isotopic number (N-Z). Here N is the number of neutrons and Z is the number of protons.
• Isobars have the same mass number and different atomic number.
  Isodiaphers are formed by successive alpha emissions and isobars are formed by successive beta emission.
  
  ${}_{82}{}^{206}\mathrm{A}$$\stackrel{-\mathrm{\alpha }}{\to }$ ${}_{80}{}^{202}\mathrm{B}$ $\stackrel{-\mathrm{\alpha }}{\to }$ ${}_{98}{}^{178}\mathrm{C}$ $\stackrel{-\mathrm{\beta }}{\to }$${}_{99}{}^{178}\mathrm{D}$ $\stackrel{-\mathrm{\beta }}{\to }$${}_{80}{}^{98}\mathrm{E}$

number of $\mathrm{\alpha }$ particles emitted successively  = 2

number of $\mathrm{\beta }$ particles emitted successively  = 2

x + y = 2 + 2 = 4