$A, B$ and $C$ are isodiaphers while $C, D$ and $E$ are isobars. Calculate the difference of protons between $A$ and $E$ .

$_{82}^{206} A ⟶ B ⟶ C ⟶ D ⟶ E .$
Given: Isodiaphers and isobars are formed in successive $α$ and $β -$ emission respectively.

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Solution

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.
$_{82}^{206} A - α - - \to_{80}^{202} B - α - - \to_{78}^{198} C - β - \to_{79}^{198} D - β - \to_{80}^{98} E$
The difference between the number of protons in A and E is $82 - 80 = 2$ .

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A,B and C are isodiaphers while C,D and E are isobars. Calculate the difference of protons between A and E. 20682A⟶B⟶C⟶D⟶E.Given: Isodiaphers and isobars are formed in successive α and β− emission respectively.

$A, B$ and $C$ are isodiaphers while $C, D$ and $E$ are isobars. Calculate the difference of protons between $A$ and $E$ .

$_{82}^{206} A ⟶ B ⟶ C ⟶ D ⟶ E .$
Given: Isodiaphers and isobars are formed in successive $α$ and $β -$ emission respectively.