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Complete and balance the following nuclear equations by supplying the missing particle

A252/98Cf + 10/5 B==========> 3 1/0n + ? (The 3 is separate from 1/0n, its not 31/0n)

b)2/1H + 3/2He ======>4/2He + ?

C1/1H + 11/5B=======>3?

d)122/53I(This is iodine) ========>122/54 Xe + ?

e)59/26Fe=======. 0/-1e + ?

Verified Answer :

A252/98Cf & 10/5 B ==> 3 1/0n & 259/103 Lr 
b) 2/1H & 3/2He ===> 4/2He & 1/1 H 
c) 1/1H & 11/5B => 3 4/2He 
d) 122/53 I ==> 122/54 Xe & 0/-1e- (Beta) 
e) 59/26Fe ==> 0/-1e & 59/27 Co

Answer: The equations are given below.

Explanation:

  • For a:

\(^{252}_{98}\textrm{Cf}+^{10}_5\textrm{B}\rightarrow ^A_Z\textrm{X}+3^{1}_{0}\textrm{n}\)

To calculate A:

Total mass on reactant side = total mass on product side

252 + 10 = A + 3

A = 259

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

98 + 5 = Z + 0

Z = 103

The isotopic symbol of lawrencium is \(_{103}^{259}\textrm{Lr}\)

  • For b:

\(^{2}_{1}\textrm{H}+^{3}_2\textrm{He}\rightarrow ^4_2\textrm{He}+^A_{Z}\textrm{X}\)

To calculate A:

Total mass on reactant side = total mass on product side

2 + 3 = A + 4

A = 1

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

1 + 2 = Z + 2

Z = 1

The isotopic symbol of hydrogen is \(_{1}^{1}\textrm{H}\)

  • For c:

\(^{1}_{1}\textrm{H}+^{11}_5\textrm{B}\rightarrow 3^A_{Z}\textrm{X}\)

To calculate A:

Total mass on reactant side = total mass on product side

1 + 11 = A

A = 12

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

1 + 5 = Z

Z = 6

The isotopic symbol of helium is \(_{2}^{4}\textrm{He}\)

  • For d:

\(^{122}_{53}\textrm{I}\rightarrow ^{122}_{54}\textrm{Xe}+_Z^A\textrm{X}\)

To calculate A:

Total mass on reactant side = total mass on product side

122 = A

A = 122

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

53 = 54 + Z  

Z = -1

The isotopic symbol of electron is \(_{-1}^{0}\textrm{e}\)

  • For e:

\(^{59}_{26}\textrm{Fe}\rightarrow ^{0}_{-1}\textrm{e}+_Z^A\textrm{X}\)

To calculate A:

Total mass on reactant side = total mass on product side

59 + 0 = A

A = 59

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

26  = -1 + Z  

Z = 27

The isotopic symbol of Cobalt is \(_{27}^{59}\textrm{Co}\)