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Balance the following nuclear reactions:...

Balance the following nuclear reactions:
a. `._(3)Li^(7) + ._(0)n^(1) rarr 2 ._(2)He^(4) + ?`
b. `._(42)Mo^(94) + ._(1)H^(2) rarr ._(0)n^(1) + ?`

Text Solution

Verified by Experts

On `LHS` of the equation
Sum of atomic number `= 3 + 0 = 3`
Sum of mass numbers `= 7 + 1 = 8`
On `RHS` of the equation,
Let the missing particles be `._(z)X^(a)`
Sum of atomic number `= Z + 4`
Sum of mass numbers `= a + 8`
Sum of atomic number of `LHS`
= Sum of atomic numbers of `RHS`
`3 = Z + 4`
or `Z = -1`
Sum of numbers of `LHS`
= Sum of mass numbers of `RHS`
`8 = a + 8`
So `a = 0`
Thus, the missing particle is `._(1)X^(0)`, i.e., `._(1)e^(0)`
The balanced equation is
`._(3)Li^(7) + ._(0)n^(1) rarr ._(2)He^(4) + ._(-1)e^(0)`
b. Lets the missing particle be `._(z)X^(a)`.
On `LHS` of the equation,
Sum of the atomic numbers `= (42 + 1) = 43`
Sum of mass numbers `= (94 + 2) = 96`
On `RHS` of the equation,
Sum of atomic numbers `= Z + 0`
Sum of mass numbers `= a + 1`
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